Answer:
(-2, 6)
Step-by-step explanation:
f(x) = x² + 4x + 10
f(x) = x² + 4x + 4 + 6
f(x) = (x + 2)² + 6
The vertex is at (-2, 6).
Todor was trying to factor 10x^2-5x+15 He found that the greatest common factor of these terms was 5 and made an area model:
Answer:
Model area of dimensions
[tex]p=5[/tex]
[tex]q=2x^2-x+3[/tex]
Step-by-step explanation:
Model for the Area
Suposse we have a rectangle of measures p and q, its area is computed by
[tex]A=p.q[/tex]
Note the expression is a product which means if an equation is found for the area, we can guess the dimensions of the supossed rectangle.
The expression for the area is
[tex]A=10x^2-5x+15[/tex]
This polynomial can be factored as
[tex]A=5(2x^2-x+3)[/tex]
We have found an explicit product, now we can guess the dimensions of the rectangle are
[tex]p=5[/tex]
[tex]q=2x^2-x+3[/tex]
Or viceversa
The grade point averages for 10 randomly selected high school students are listed below. Assume the grade point averages are normally distributed. 2.0 3.2 1.8 2.9 0.9 4.0 3.3 2.9 3.6 0.8 Find a 98% confidence interval for the true mean.
Answer:
[tex]2.54-2.82\frac{1.110}{\sqrt{10}}=1.55[/tex]
[tex]2.54+2.82\frac{1.110}{\sqrt{10}}=3.53[/tex]
So on this case the 98% confidence interval would be given by (1.55;3.53)
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
[tex]\bar X[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
s represent the sample standard deviation
n represent the sample size
Solution to the problem
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
In order to calculate the mean and the sample deviation we can use the following formulas:
[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)
[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)
The mean calculated for this case is [tex]\bar X=2.54[/tex]
The sample deviation calculated [tex]s=1.110[/tex]
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=10-1=9[/tex]
Since the Confidence is 0.98 or 98%, the value of [tex]\alpha=0.02[/tex] and [tex]\alpha/2 =0.01[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.01,9)".And we see that [tex]t_{\alpha/2}=2.82[/tex]
Now we have everything in order to replace into formula (1):
[tex]2.54-2.82\frac{1.110}{\sqrt{10}}=1.55[/tex]
[tex]2.54+2.82\frac{1.110}{\sqrt{10}}=3.53[/tex]
So on this case the 98% confidence interval would be given by (1.55;3.53)
The 98% confidence interval would be given by (1.55;3.53)
A range of values that, with a particular level of confidence, is likely to encompass a population value is called a confidence interval. A population mean is typically stated as a percentage that falls between an upper and lower interval.
The range of values in a confidence interval below and above the sample statistic is called the margin of error.
The normal distribution is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
show the sample mean for the given sample.
population mean (the relevant variable)
The sample standard deviation is denoted by s.
n stands for the number of samples.
Resolution of the issue
The following formula produces the mean's confidence interval:
[tex]\bar x[/tex] ± [tex]\frac{t_a}{2} \frac{s}{\sqrt{n} }[/tex] ____________(1)
In order to calculate the mean and the sample deviation we can use the following formulas:
[tex]\bar x =[/tex] [tex]\sum_{i =1}^n \frac{x_i}{n}[/tex]_______(2)
[tex]s = \sqrt{\frac{\sum^n_{i=1}(x_i-\bar x)}{n-1} }[/tex] ____________(3)
The mean calculated for this case is X = 2.54
The sample deviation calculated s = 1.110
t In order to calculate the critical value [tex]\frac{t_a}{2}[/tex] we need freedom, given by: to find first the degrees of
df = n-1=10-19
Since the Confidence is 0.98 or 98%, the value of a = 0.02 and a/2 = 0.01, and we can use excel, a calculator or a tabel to find the critical value.
[tex]\frac{t_a}{2}[/tex] = 2.82
Now we have everything in order to replace into formula (1):
2.54 - 2.82 [tex]\frac{1.110}{\sqrt{10} }[/tex] = 1.55
2.54 + 2.82 [tex]\frac{1.110}{\sqrt{10} }[/tex] = 3.53
A brochure claims that the average maximum height for a certain type of plant is 0.7 m. A gardener suspects that this is not accurate locally due to variation in soil conditions, and believes the local height is shorter. A random sample of 40 mature plants is taken. The mean height of the sample is 0.65 m with a standard deviation of 0.20 m. Test the claim that the local mean height is less than 0.7 m using a 5% level of significance.
Answer:
As [tex]Z<-Z_{\alpha}[/tex], it is possible to reject null hypotesis. It means that the local mean height is less tha 0.7 m with a 5% level of significance.
Step-by-step explanation:
1. Relevant data:
[tex]\mu=0.70\\N=40\\\alpha=0.05\\X=0.65\\s=0.20[/tex]
2. Hypotesis testing
[tex]H_{0}=\mu=0.70[/tex]
[tex]H_{1} =\mu< 0.70[/tex]
3. Find the rejection area
From the one tail standard normal chart, whe have Z-value for [tex]\alpha=0.05[/tex] is 1.56
Then rejection area is left 1.56 in normal curve.
4. Find the test statistic:
[tex]Z=\frac{X-\mu_{0} }{\sigma/\sqrt{n}}[/tex]
[tex]Z=\frac{0.65-0.70}{0.20/\sqrt{40}}\\Z=-1.58[/tex]
5. Hypotesis Testing
[tex]Z_{\alpha}=1.56\\Z=-1.58[/tex]
[tex]-1.58<-1.56[/tex]
As [tex]Z<-Z_{\alpha}[/tex], it is possible to reject null hypotesis. It means that the local mean height is less tha 0.7 m with a 5% level of significance.
Please help!! It’s due @ midnight
How much should Marc deposit weekly into an account at 8% compounded weekly in order to have
$4500 available for a round trip plane ticket, hotel, and spending money for a trip to Sweden in 2 years?
Please give step by step!
Answer:
ooh i just learned this, not 100% sure but the amount he should have to deposit is $3835.12
if the weekly one means depositing money every week then it would be $36.88 I think.
Step-by-step explanation:
p=?
r=.08
n=52
t=2
4500 = P(1+.08/52)^(52 x 2)
divide both sides by (1+.08/52)^(52 x 2)
and you are left with $3835.12
if i take into account that Marc is depositing the money every week the i would divide it by 104 (that is 52 x 2 because 52 weeks in a year and it says 2 years) you would be left with $36.88.
Hope I was any help.
Answer: Marc should deposit $39.87 weekly.
Step-by-step explanation:
We would apply the formula for determining future value involving deposits at constant intervals. It is expressed as
S = R[{(1 + r)^n - 1)}/r][1 + r]
Where
S represents the future value of the investment.
R represents the regular payments made(could be weekly, monthly)
r = represents interest rate/number of interval payments.
n represents the total number of payments made.
From the information given,
S = $4500
Assuming there are 52 weeks in a year, then
r = 0.08/52 = 0.0015
n = 52 × 2 = 104
Therefore,
4500 = R[{(1 + 0.0015)^104 - 1)}/0.0015][1 + 0.0015]
4500 = R[{(1.0015)^104 - 1)}/0.0015][1.0015]
4500 = R[{(1.169 - 1)}/0.0015][1.0015]
4500 = R[{(0.169)}/0.0015][1.0015]
4500 = R[112.67][1.0015]
4500 = 112.839R
R = 4500/112.839
R = 39.87
A poll showed that 57.4% of Americans say they believe that statistics teachers know the true meaning of life. What is the probability of randomly selecting someone who does not believe that statistics teachers know the true meaning of life. Report your answer as a decimal accurate to 3 decimal places.
Answer:
The probability of randomly selecting someone who does not believe that statistics teachers know the true meaning of life is 0.426.
Step-by-step explanation:
Let X = number of Americans who believe that statistics teachers know the true meaning of life.
The probability of the random variable X is,
P (X) = 0.574
The event of a person not believing that statistics teachers know the true meaning of life is the complement of the event X.
The probability of the complement of an event, E is the probability of that event not happening.
[tex]P(E^{c})=1-P(E)[/tex]
Compute the value of [tex]P(X^{c})[/tex] as follows:
[tex]P(X^{c})=1-P(X)\\=1-0.574\\=0.426[/tex]
Thus, the probability of randomly selecting someone who does not believe that statistics teachers know the true meaning of life is 0.426.
Answer:
Probability of randomly selecting someone who does not believe that statistics teachers know the true meaning of life = 0.426 .
Step-by-step explanation:
We are given that a poll showed that 57.4% of Americans say they believe that statistics teachers know the true meaning of life.
Let the above probability that % of Americans who believe that statistics teachers know the true meaning of life = P(A) = 0.574
Now, probability of randomly selecting someone who does not believe that statistics teachers know the true meaning of life is given by = 1 - Probability of randomly selecting someone who believe that statistics teachers know the true meaning of life = 1 - P(A)
So, required probability = 1 - 0.574 = 0.426 .
Let y be a random variable with a known distribution, and consider the square loss function `(a; y) = (a????y)2. We want to find the action a that has minimal risk, namely, to find a = arg mina E(a ???? y)2, where the expectation is with respect to y. Show that a = Ey, and the Bayes risk (i.e. the risk of a) is Var(y). In other words, if you want to try to predict the value of a random variable, the best you can do (for minimizing expected square loss) is to predict the mean of the distribution. Your expected loss for predicting the mean will be the variance of the distribution. You should use the fact that Var(y) = Ey2 ???? (Ey)2.
Answer/ Explanation:
Since X is exponentially distributed, its expected value is given by E[X]=1/λ=2.
Therefore, E[Y]=E[1−2X]=E[1]+E[−2X]=E[1]−2E[X]=1−2E[X]=1−2⋅2=−3.
Hence,
We define the moment-generating function of Y as MY(t). It is given by
MY(t)=E[etY]=E[et(1−2X)]=E[ete−2tX]=E[et]E[e−2tX].
If I give you the hint that E[g(Y)]=∫∞0g(y)fY(y)dy, where fY(y) is the probability density function of Y, can you also solve for the moment generating function of Y?
We have E[X2]=2/λ2=2/(0.5)2=8. Thus,
E[Y2]=E[(1−2X)2]=E[1−4X+4X2]=E[1]−4E[X]+4E[X2]=1−4⋅2+4⋅8=25.
So,
Var(Y)=E[Y2]−E[Y]2=25−(−3)2=16.
Continuing for the moment-generating function:
MY(t)=E[et]E[e−2tX]=etE[e−2tX]=et∫∞x=0e−2txfX(x)dx,
where fX(x) is the probability density function of X and thus satisfies fX(x)=λe−λx. Substituting yields
MY(t)=et∫∞x=0e−2txλe−λxdx=λet∫∞x=0e−x(2t+λ)dx=λet2t+λ.
It is also good to note that
If you are after expectation, variance or moment generating function of Y then it is not needed to find the PDF of Y (see the answer of Ritz).
This is not an answer on the question in the title, but one on the question in the body.
FY(y)=P(Y≤y)=P(1−2X≤y)=P(X≥0.5−0.5y)=1−FX(0.5−0.5y)
Note that the last equality demands that FX is continuous.
Differentating on both sides gives fY on LHS and an expression in fX on RHS.
Two hikers come to a ravine and want to know how wide it is. They set up two similar triangles as shown in the diagram. How far is it across the ravine?
Answer:
d = 80 ft
Step-by-step explanation:
Start with angle BCA, tan(BCA) = 30/15 so BCA = 63.43 degrees.
BCA = ECD since they are opposite angles.
tan ECD = d/40
40 tan(63.4degrees) = d
d = 80 ft
Alternatively, you can use similar triangles since the angles are the same, BCA = ECD, CED = CAB, and CBA = EDC. In that case use proportions to get 30/15 = d/40 so 2 = d/40 and d = 80.
A golfer's bag contains 24 golf balls, 18 of which are ProFlight brand and the other 6 are DistMax brand. Find the probability that he reaches in his bag and randomly selects 5 golf balls and 4 of them are ProFlights and the other 1 is DistMax.
Answer:
1. Assuming with replacement, the probability is 7.91%; or
2. Assuming without replacement, the probability is 8.64%
Step-by-step explanation:
Total number of golf balls = 24
Let Pr denotes probability, P denotes ProFlights, D denotes DistMax.
The probability of selecting 5 balls can be with or without replacement. Since the question is silent on this, the answers to the methods are provided as follows:
1. Assuming with replacement
Pr(4 P and 1 D) = (18/24) × (18/24) × (18/24) × (18/24) × (6/24)
= 0.75 × 0.75 × 0.75 × 0.75 × 0.25
= 0.0791 = 7.91%
2. Assuming without replacement
Here, we assume that 4 ProFlights are selected first before 1 DistMax is selected, and the probability is as follows:
Pr(4 P and 1 D) = (18/24) × (17/23) × (16/22) × (15/21) × (6/20)
= 0.7500 × 0.7391 × 0.7273 × 0.7143 × 0.3000
= 0.0864 = 8.64%
Therefore, the probability that he reaches in his bag and randomly selects 5 golf balls and 4 of them are ProFlights and the other 1 is DistMax is 7.91% assuming with replacement or 8.64% assuming without replacement.
Final answer:
To find the probability, we use the concept of combinations.
Explanation:
To find the probability that the golfer randomly selects 5 golf balls and 4 of them are ProFlights and the other 1 is DistMax, we need to use the concept of combinations. The total number of ways to select 5 golf balls from a bag containing 24 balls is C(24,5). The number of ways to select 4 ProFlight balls and 1 DistMax ball from their respective totals is C(18,4) * C(6,1). Therefore, the probability is given by:
P = (C(18,4) * C(6,1)) / C(24,5)
Find ∂w/∂s and ∂w/∂t using the appropriate Chain Rule.
Function Point
w = y3 − 9x2y
x = es, y = et
s = −5, t = 10
Evaluate each partial derivative at the given values of s and t.
Answer:
The value of [tex]\frac{\partial w}{\partial s}[/tex] is [tex]e^{3t}-18e^{2s+t}[/tex]The value of [tex]\frac{\partial w}{\partial t}[/tex] is [tex]3e^{3t}-9e^{2s+t}[/tex]The partial derivative at s=-5 and t=10 is [tex]\frac{\partial w}{\partial s}[/tex] is [tex]e^{30}-18[/tex]The partial derivative at s=-5 and t=10 is [tex]\frac{\partial w}{\partial t}=3(e^{30}-3)[/tex]Step-by-step explanation:
Given that the Function point are [tex]w=y^3-9x^2y[/tex]
[tex]x=e^s[/tex], [tex]y=e^t[/tex] and s = -5, t = 10
To find [tex]\frac{\partial w}{\partial s}[/tex] and [tex]\frac{\partial w}{\partial t}[/tex]using the appropriate Chain Rule :[tex]w=y^3-9x^2y[/tex]
Substitute the values of x and y in the above equation we get
[tex]w=(e^t)^3-9(e^s)^2(e^t)[/tex]
[tex]w=e^{3t}-9e^{2s}.e^t[/tex]
Now partially differentiating w with respect to s by using chain rule we have[tex]\frac{\partial w}{\partial ∂s}=e^{3t}-9(e^{2s}).2(e^t)[/tex]
[tex]=e^{3t}-18e^{2s}.(e^t)[/tex]
[tex]=e^{3t}-18e^{2s+t}[/tex]
Therefore the value of [tex]\frac{\partial w}{\partial s}[/tex] is [tex]e^{3t}-18e^{2s+t}[/tex][tex]w=e^{3t}-9e^{2s}.e^t[/tex]
Now partially differentiating w with respect to t by using chain rule we have[tex]\frac{\partial w}{\partial t}=e^{3t}.(3)-9e^{2s}(e^t).(1)[/tex]
[tex]=3e^{3t}-9e^{2s+t}[/tex]
Therefore the value of [tex]\frac{\partial w}{\partial t}[/tex] is [tex]3e^{3t}-9e^{2s+t}[/tex]Now put s-5 and t=10 to evaluate each partial derivative at the given values of s and t :
[tex]\frac{\partial w}{\partial s}=e^{3t}-18e^{2s+t}[/tex]
[tex]=e^{3(10}-18e^{2(-5)+10}[/tex]
[tex]=e^{30}-18e^{-10+10}[/tex]
[tex]=e^{30}-18e^0[/tex]
[tex]=e^{30}-18[/tex]
Therefore the partial derivative at s=-5 and t=10 is [tex]\frac{\partial w}{\partial s}[/tex] is [tex]e^{30}-18[/tex][tex]\frac{\partial w}{\partial t}=3e^{3t}-9e^{2s+t}[/tex]
[tex]=3e^{3(10)}-9e^{2(-5)+10}[/tex]
[tex]=3e^{30}-9e{-10+10}[/tex]
[tex]=3e^{30}-9e{0}[/tex]
[tex]=3e^{30}-9[/tex]
[tex]\frac{\partial w}{\partial t}=3(e^{30}-3)[/tex]
Therefore the partial derivative at s=-5 and t=10 is [tex]\frac{\partial w}{\partial t}=3(e^{30}-3)[/tex]Using the Chain Rule, it is found that:
[tex]\frac{\partial W}{\partial s}(-5,10) = -18[/tex]
[tex]\frac{\partial W}{\partial s}(-5,10) = 3e^{30} - 9[/tex]
w is a function of x and y, which are functions of s and t, thus, by the Chain Rule:
[tex]\frac{\partial W}{\partial s} = \frac{\partial W}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial W}{\partial y}\frac{\partial y}{\partial s}[/tex]
[tex]\frac{\partial W}{\partial t} = \frac{\partial W}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial W}{\partial y}\frac{\partial y}{\partial t}[/tex]
Then, the derivatives are:
[tex]\frac{\partial W}{\partial x} = -18xy[/tex]
[tex]\frac{\partial x}{\partial s} = e^s[/tex]
[tex]\frac{\partial W}{\partial y} = 3y^2 - 9x^2[/tex]
[tex]\frac{\partial y}{\partial s} = 0[/tex]
Then
[tex]\frac{\partial W}{\partial s} = \frac{\partial W}{\partial x}\frac{\partial x}{\partial s}[/tex]
[tex]\frac{\partial W}{\partial s} = -18xy(e^s)[/tex]
[tex]\frac{\partial W}{\partial s} = -18e^se^t(e^s)[/tex]
[tex]\frac{\partial W}{\partial s} = -18e^{2s}e^t[/tex]
[tex]\frac{\partial W}{\partial s} = -18e^{2s + t}[/tex]
At s = -5 and t = 10:
[tex]\frac{\partial W}{\partial s}(-5,10) = -18e^{2(-5) + 10} = -18[/tex]
Then, relative to t:
[tex]\frac{\partial x}{\partial t} = 0[/tex]
[tex]\frac{\partial y}{\partial t} = e^t[/tex]
[tex]\frac{\partial W}{\partial t} = \frac{\partial W}{\partial y}\frac{\partial y}{\partial t}[/tex]
[tex]\frac{\partial W}{\partial t} = (3y^2 - 9x^2)e^t[/tex]
[tex]\frac{\partial W}{\partial t} = (3e^{2t} - 9e^{2s})e^t[/tex]
At s = -5 and t = 10:
[tex]\frac{\partial W}{\partial s}(-5,10) = (3e^{20} - 9e^{-10})e^{10} = 3e^{30} - 9[/tex]
A similar problem is given at https://brainly.com/question/10309252
The method of Lagrange multipliers assumes that the extreme values exist, but that is not always the case. Show that the problem of finding the minimum value of f(x, y) = x^2 + y^2 subject to the constraint can be solved using Lagrange multipliers, but does not have a maximum value with that constraint.
Answer:
Incomplete question check attachment for complete question
Step-by-step explanation:
Given the function,
F(x, y)=x²+y²
The La Grange is theorem
Solve the following system of equations.
∇f(x, y)= λ∇g(x, y)
g(x, y)=k
Fx=λgx
Fy=λgy
Fz=λgz
Plug in all solutions, (x,y), from the first step into f(x, y) and identify the minimum and maximum values, provided they exist and
∇g≠0 at the point.
The constant, λ, is called the Lagrange Multiplier.
F(x, y)=x²+y²
∇f= 2x i + 2y j
So, given the constraint is xy=1.
g(x, y)= xy-1=0
∇g= y i + x j
gx= y. And gy=x
So, here is the system of equations that we need to solve.
Fx=λgx; 2x=λy. Equation 1
Fy=λgy; 2y=λx. Equation 2
xy=1
Solving this
x=λy/2. From equation 1, now substitute this into equation 2
2y=λ(λy/2)
2y=λ²y/2
2y-λ²y/2 =0
y(2-λ²/2)=0
Then, y=0. Or (2-λ²/2)=0
-λ²/2=-2
λ²=4
Then, λ= ±2
So substitute λ=±2 into equation 2
2y=2x
Then, y=x
So inserting this into the constraint g will give
xy=1. Since y=x
x²=1
Therefore,
x=√1
x=±1
Also y=x
Then, y=±1
Therefore, there are four points that solve the system above.
(1,1) (-1,-1) (1,-1) and (-1,1)
The first two points (1,1) (-1,-1) shows the minimum points because they show xy=1
The other points does not give xy=1
They give xy=-1.
Now,
F(x, y)=x²+y²
F(1,1)=1²+1²
F(1,1)=2
F(-1,-1)= (-1) ²+(-1)²
F(-1,-1)=1+1
F(-1,-1)=2
Then
F(1,1)= F(-1,-1)=2 is the minimum point
This gives the same four points as we found using Lagrange multipliers above.
The owner of a fish market determined that the mean weight for a catfish is 3.2 pounds with a standard deviation of 0.8 pound. Assuming the weights of catfish are normally distributed, the probability that a randomly selected catfish will weigh between 3 and 5.4 pounds is ________.
Answer: The probability that a randomly selected catfish will weigh between 3 and 5.4 pounds is 0.596
Step-by-step explanation:
Since the weights of catfish are assumed to be normally distributed,
we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = weights of catfish.
µ = mean weight
σ = standard deviation
From the information given,
µ = 3.2 pounds
σ = 0.8 pound
The probability that a randomly selected catfish will weigh between 3 and 5.4 pounds is is expressed as
P(x ≤ 3 ≤ 5.4)
For x = 3
z = (3 - 3.2)/0.8 = - 0.25
Looking at the normal distribution table, the probability corresponding to the z score is 0.401
For x = 5.4
z = (5.4 - 3.2)/0.8 = 2.75
Looking at the normal distribution table, the probability corresponding to the z score is 0.997
Therefore,.
P(x ≤ 3 ≤ 5.4) = 0.997 - 0.401 = 0.596
To determine the probability that a catfish will weigh between 3 and 5.4 pounds, we calculate the z-scores for these weights and find the corresponding probabilities. The probability is approximately 0.5957 or 59.57%.
Explanation:The probability that a randomly selected catfish will weigh between 3 and 5.4 pounds can be calculated using the standard normal distribution.
First, we convert the weights to z-scores using the formula:
Z = (X - μ) / σ
For 3 pounds:
Z = (3 - 3.2) / 0.8 = -0.25
For 5.4 pounds:
Z = (5.4 - 3.2) / 0.8 = 2.75
Next, we look up these z-scores on the z-table or use a calculator with normal distribution functions to find the probabilities.
P(Z < 2.75) = 0.9970 (rounded to four decimal places)
P(Z < -0.25) = 0.4013 (rounded to four decimal places)
Then we find the difference to determine the probability of a catfish weighing between these two values:
Probability = P(Z < 2.75) - P(Z < -0.25)
Probability = 0.9970 - 0.4013 = 0.5957
The probability that a randomly selected catfish will weigh between 3 and 5.4 pounds is approximately 0.5957 or 59.57%.
The level of nitrogen oxides (NOX) in a exhaust of cars of a particular model varies normally with mean 0.25 grams per miles and standard deviation 0.05 g/mi. government regulations call for NOX emissions no higher than 0.3 g/mi.
a. What is the probability that a single car of this model fails to meet the NOX requirement?
b. A company has 4 cars of this model in its fleet. What is the probability that the average NOX level of these cars are above 0.3 g/mi limit?
Answer:
a) 15.87% probability that a single car of this model fails to meet the NOX requirement.
b) 2.28% probability that the average NOX level of these cars are above 0.3 g/mi limit
Step-by-step explanation:
We use the normal probability distribution and the central limit theorem to solve this question.
Normal probability distribution:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central limit theorem:
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 0.25, \sigma = 0.05[/tex]
a. What is the probability that a single car of this model fails to meet the NOX requirement?
Emissions higher than 0.3, which is 1 subtracted by the pvalue of Z when X = 0.3. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.3 - 0.25}{0.05}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a pvalue of 0.8417.
1 - 0.8413 = 0.1587.
15.87% probability that a single car of this model fails to meet the NOX requirement.
b. A company has 4 cars of this model in its fleet. What is the probability that the average NOX level of these cars are above 0.3 g/mi limit?
Now we have [tex]n = 4, s = \frac{0.05}{\sqrt{4}} = 0.025[/tex]
The probability is 1 subtracted by the pvalue of Z when X = 0.3. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.3 - 0.25}{0.025}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a pvalue of 0.9772
1 - 0.9772 = 0.0228
2.28% probability that the average NOX level of these cars are above 0.3 g/mi limit
The probability that a single car of this model fails to meet the NOX requirement is 15.87%.
What is z score?Z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:
z = (raw score - mean) / standard deviation
Given; mean of 0.25 g and a standard deviation of 0.05 g/mi
a) For > 0.3:
z = (0.3 - 0.25)/0.05 = 1
P(z > 1) = 1 - P(z < 1) = 1 - 0.8413 = 0.1587
b) For > 0.3, sample size = 4
z = (0.3 - 0.25)/(0.05 ÷√4) = 2
P(z > 2) = 1 - P(z < 2) = 1 - 0.9772 = 0.0228
The probability that a single car of this model fails to meet the NOX requirement is 15.87%.
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Let X = the time (in 10−1 weeks) from shipment of a defective product until the customer returns the product. Suppose that the minimum return time is γ = 3.5 and that the excess X − 3.5 over the minimum has a Weibull distribution with parameters α = 2 and β = 1.5.
(a) What is the cdf of X?
F(x) = 0 x < 3.5
1−e^−((x−3.5)2.5)2 x ≥ 3.5
(b) What are the expected return time and variance of return time? [Hint: First obtain
E(X − 3.5)
and
V(X − 3.5).]
(Round your answers to three decimal places.)
E(X) = 10^−1 weeks
V(X) = (10^−1 weeks)2
(c) Compute
P(X > 6).
(Round your answer to four decimal places.)
Answer:
Step-by-step explanation: see attachment for solution
If X is the time in 10 to 1 week of the shipment of a defective product until the customer returns the product.
Now suppose the mini returns y = 3.5 and the excess X is 3.5 over the mini has a Weibull distribution with parameters Alfa 2 and Beta 1.5.The shipment of the defective product will be
A. 1-e x(25)2. B. 5/4 .Expected value is E (X) = 5.7125.
C. P(X>5) = 0.3679.Learn more about the shipment of a defective product
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Which expression can be simplified as 625/n^12 check all that apply
Answer:
Only options, A and E give 625/n¹² on simplification. The other options do not apply.
(5n⁻³)⁴ = (625/n¹²)
(25n⁻⁶)² = (625/n¹²)
Step-by-step explanation:
625/n¹²
a) (5n⁻³)⁴
According to the law of indices, this becomes
(5⁴)(n⁻³)⁴ = 625(n⁻¹²) = 625/n¹²
This applies!
b) (5n⁻³)⁻⁴
According to the law of indices, this becomes
(5⁻⁴)(n⁻³)⁻⁴ = (n¹²)/625 = n¹²/625
Does Not apply!
c) (5n⁻⁴)³
This becomes
(5³)(n⁻⁴)³ = 125n⁻¹² = 125/n¹²
Does Not apply!
d) (25n⁻⁶)⁻²
This becomes
(25⁻²)(n⁻⁶)⁻² = n⁻¹²/625 = 1/(625n¹²)
Does Not apply!
e) (25n⁻⁶)²
This becomes
(25²)(n⁻⁶)² = 625n⁻¹² = 625/n¹²
This applies!
Answer:
A AND E
Step-by-step explanation:
16. A car depreciates in value at a rate of 10%. The car currently has a value of $12,000.
What will be its value in 10 years?
Answer: $4,184.14
Step-by-step explanation:
If a car depreciates, it mean the the car is losing it's marketable worth and sometimes at a constant rate. The worth of the car after some years does not remain the same.
The formula for this depreciation when at a constant rate is denoted as:
D = P × [1 - r/100]^n
Where:
D=the Depreciated value of the car after n period which is what is being determined.
P = initial value of the car before depreciation is considered and in this case, P = $12,000
r = constant Rate Of depreciation and in this case = 10%
n = period being considered, which in this case, n = 10years.
Hence,
D = 12,000 × [1 - 10/100]^10
D = $4,184.14
Simplify.
25
−
40
A) 3
10
B) 10
10
C) 5 − 2
10
D) 2
10
− 5
Answer:
5 -2√10 = C
Step-by-step explanation:
√25 -√40
√25 = √5 x5 = √5² = 5
√40 = √5 x 8 = √5x 2x4 = √10x4 = √10x 2² = 2√10 (when the 2 comes back into the square root it become 2²)
√25 -√40 = 5 -2√10
Some scientists believe that the average surface temperature of the world has been rising steadily. The average surface temperature can be modeled by T = 0.02t + 15.0. where T is temperature in ∘C and t is years since 1950.
(a) What do the slope and T-intercept represent?
(b) Use the equation to predict the average global surface temperature in 2050.
Answer:
(a)
The slope of the equation = 0.02
Therefore T-intercept equals 15
(b)
Therefore the average global surface temperature in 2050 is 17°C.
Step-by-step explanation:
If a equation is in the form
y= mx+c........(1)
Then the slope of the equation is m.
Slope: The tangent of the angle which is made with the positive x-axis.
If θ be the angle , Then slope (m)= tanθ.
If a equation in the form
[tex]\frac{x}{a} +\frac{y}{b} =1[/tex]............(2)
Then x-axis intercept equals a and y-axis intercept equals b.
Given equation is
T=0.02t+15.0
Comparing with equation (1)
The slope of the equation = 0.02
Again we can rewrite the equation as
[tex]T-0.02t=15[/tex]
[tex]\Rightarrow \frac{T}{15} -\frac{0.02t}{15}=1[/tex]
Comparing with (2)
Therefore T-intercept equals 15
(b)
Here t= 2050-1950 =100
Putting t=100 in the given equation
T=0.02(100)+15 = 2+15 =17
Therefore the average global surface temperature in 2050 is 17°C.
Solve using normalcdf
Let X be the random variable representing monthly trainee income. X is distributed with mean $1100 and standard deviation $150. You want to find the proportion of trainees that earn less than $900 per month, or Pr(X < 900).
Using normalcdf (on a TI calculator, for instance), you would compute
normalcdf(-1E99, 900, 1100, 150)
to get a proportion of approximately 0.09121, or 9.121%.
That is, the syntax for normalcdf is
normalcdf(lower limit, upper limit, mean, standard deviation)
In this case, you pick a very large negative number for "lower limit" in order to simulate negative infinity.
In a study of the effects of acid rain, a random sample of 100 trees from a particular forest is examined. Forty percent of these show some signs of damage. Which of the following statements is correct?
a.None of the above
b. The sampling distribution of the proportion of damaged trees is approximately Normal
c.This is a comparative experiment.
d.If a sample of 1000 trees was examined, the variability of the sample proportion would be larger than in a sample of 100 trees
e.If we took another random sample of trees, we would find that 40% of these would show some signs of damage
Answer: B
Step-by-step explanation:
The sampling distribution of the proportion of damaged trees is approximately Normal
Answer:B. The sampling distribution of the proportion of damaged trees is approximately Normal.
Step-by-step explanation: Sampling distribution is a probability distribution of a statistic which is gotten through the collection of a large number of samples from the population of interest.
A normal distribution is a distribution of the samples symmetrically around the mean, when a Statistical metric shows that the outcome is distributed around the central region it shows that the distribution is normal.
FOR A FORTY PERCENT INCIDENCE, THE SAMPLING DISTRIBUTION OF THE PROPORTION OF DAMAGED TREES IS APPROXIMATELY NORMAL AS IT IS CLOSE TO 50% WHICH IS THE MEAN OF 100%.
In human resource management, performance of employees is measured as a numerical score which is assumed to be normally distributed. The mean score is 150 and the standard deviation 13. What is the probability that a randomly selected employee will have a score less than 120?
Answer:
[tex]P(X<120)=P(\frac{X-\mu}{\sigma}<\frac{120-\mu}{\sigma})=P(Z<\frac{120-150}{13})=P(z<-2.308)[/tex]
And we can find this probability using the normal standard table or excel and we got:
[tex]P(z<-2.308)=0.0105[/tex]
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(150,13)[/tex]
Where [tex]\mu=150[/tex] and [tex]\sigma=13[/tex]
We are interested on this probability
[tex]P(X<120)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X<120)=P(\frac{X-\mu}{\sigma}<\frac{120-\mu}{\sigma})=P(Z<\frac{120-150}{13})=P(z<-2.308)[/tex]
And we can find this probability using the normal standard table or excel and we got:
[tex]P(z<-2.308)=0.0105[/tex]
1. Write an equivalent expression for 27x+18
2. Write the inequality this number line represents
3.erin is going to paint a wall in her house she needs to find the area of the wall so she knows how much paint to purchase what is the area of her wall
4.walt received a package that is 12 1/3 inches long 6 3/4 inches high and 8 1/2 inches wide what is the surface area of the package
Answer:
1. 27x+18 = x+x+x+x+x......+x + 18
You sum "x" 27 times.
2. [tex](36,\infty)[/tex]
3. [tex]285/2 = 142.4[/tex]
4. 2*(12 1/3 )*(8 1/2) + 2*(12 1/3 )*(6 3/4)+2*(6 3/4 )*(8 1/2)
Step-by-step explanation:
1. Remember that multiplication is a simplification of the sum, so, when you say for example, 4*3, that actually means 3+3+3+3, similarly, when you say, 27x, that means x+x+x...+x 27 times.
2. From the image you can see that
The 36 is NOT taken, and then you go all the way to infinity, therefore we say [tex](36,\infty)[/tex]. Suppose that 36 was taken, then we would say [tex][36,\infty)[/tex].
3. From the attached photo
you can see that we can compute first the area of the rectangle with length = 15 and height = 7, and also note that at the top a triangle with base 15 and height 5 is formed, so the area of the whole figure would be the area of the rectangle at the bottom plus the area of the triangle on top. That would be 7*15+(15*5)/2 = 285/2
4. Remember that in general the formula for surface area would be
[tex]2lw +2lh+2wh[/tex]
Where l = length , w = wide, h = height. In this case l = 12 1/3 , w = 8 1/2 and h = 6 3/4
Determine the original set of data. 1 0 1 5 2 1 4 4 7 9 3 3 5 5 5 7 9 4 0 1 Legend: 1|0 represents 10The originat set of the data is?
The data set is S = {10, 11, 15, 21, 24, 24, 27, 29, 33, 35, 35, 37, 39, 40, 40}
A stem-and-leaf plot is a method to represent the data in tabular form.
The stem consist of the first digits of the data values arranged in ascending order.
The leaf consist of the remaining digits.
The data provided is:
Stem | Leaf
1 | 0 1 5
2 | 1 4 4 7 9
3 | 3 5 5 7 9
4 | 0 1
The original data is:
10, 11, 15, 21, 24, 24, 27, 29, 33, 35, 35, 37, 39, 40, 40
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The original set of data consists of number sequences as follows: 10 12 4 7 11 4 3 10 0, 10 4 14 11 13 2 4 6, 12 6 9 10, 5 13 4, 10 14 12 11, and 6 10 11 0 11 13 2.
The original set of data is:
10 12 4 7 11 4 3 10 010 4 14 11 13 2 4 612 6 9 105 13 410 14 12 116 10 11 0 11 13 2Consider the four points used in Problem 1: (-9,13), (-3, 9), (3, 6) and (9, 1) Three out of four of these points lie on one line. Which one of the four points does not lie on the same line as the other three? Justify your answer using slope. b) Find the equation of the line that contains three out of the four points
Answer:
a) ( 3 , 6 )
b) y = -2*x / 3 + 7
Step-by-step explanation:
Given:
- The four points are:
(-9,13), (-3, 9), (3, 6) and (9, 1)
- Three points lie on the same line.
Find:
Which one of the four points does not lie on the same line as the other three?
Solution:
- Compute the slope between each pair of point:
(-9,13), (-3, 9)
m_1 = ( 9 - 13 ) / ( -3 + 9 ) = - 2/3
(-9,13), (3, 6)
m_2 = ( 6 - 13 ) / ( 3 + 9 ) = - 0.5833
(-9,13), (9, 1)
m_3 = ( 1 - 13 ) / ( 9 + 9 ) = - 2/3
- We see that m_1 = m_3 ≠ m_2 . Hence, point ( 3 ,6 ) does not lie on the same line.
- The equation of line is expressed as:
y = m*x + C
m = -2/3
y = -2*x / 3 + C
1 = -2*9 / 3 + C ....... ( 9 , 1 )
C = 7
- The equation of the line is:
y = -2*x / 3 + 7
Points A(-9, 13), B(-3, 9) and D(9, 1) are colinear.
b). Equation of the line passing through these points will be [tex]y=-\frac{2}{3}x+7[/tex]
Property for the colinear points:If three points are colinear, slope of the lines joining these points will be equal.Slope of a line joining two points [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex] is given by,
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
Given points in the question,
A(-9, 13), B(-3, 9), C(3, 6) and D(9, 1)
If the slopes of the lines joining these points are same, points will be colinear.
Slope of AB = [tex]\frac{13-9}{-9+3}=-\frac{2}{3}[/tex]
Slope of AC = [tex]\frac{13-6}{-9-3}=-\frac{7}{12}[/tex]
Slope of AD = [tex]\frac{13-1}{-9-9}=-\frac{2}{3}[/tex]
Since, slopes of AB and AD are equal, points A, B and D will be colinear.
b). Let the equation of the line passing through A, B and D is,
y - y' = m(x - x')
Since, this line passes through D(9, 1) and slope = [tex]-\frac{2}{3}[/tex]
Equation of the line → [tex]y-1=-\frac{2}{3}(x-9)[/tex]
[tex]y=-\frac{2}{3}x+6+1[/tex]
[tex]y=-\frac{2}{3}x+7[/tex]
Therefore, points A(-9, 13), B(-3, 9) and D(9, 1) are colinear and equation of the line passing through these points will be [tex]y=-\frac{2}{3}x+7[/tex].
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A study of Machiavellian traits in lawyers was performed. Machiavellian describes negative character traits such as manipulation, cunning, duplicity, deception, and bad faith. A Mach rating score was determined for each in a sample of lawyers. The lawyers were then classified as having high comma moderate comma or low Mach rating scores. The researcher investigated the impact of both Mach score classification and gender on the average income of a lawyer. For this experiment, identify Bold a. the experimental unit, Bold b. the response variable, Bold c. the factors, Bold d. the levels of each factor, and Bold e. the treatments.
a. The experimental unit:
The experimental units would be the lawyers that participate in this experiment. The experimental units are the subjects upon which the experiment is performed.
b. The response variable:
The response variable would be income. This is the variable that measures the response or outcome of the study.
c. The factors:
The factors are the variables whose levels are manipulated by the researcher. In this case, these would be the Mach score, classification and gender.
d. The levels of each factor:
The levels would include the levels of the Mach score (high, moderate, low) and the levels of gender (male, female).
e. The treatments:
The treatments are all the possible combinations of one level of each factor. Therefore, these are: High and male, high and female, moderate and men, moderate and female, low and male, low and female.
Suppose Team A has a 0.75 probability to win their next game and Team B has a 0.85 probability to win their next game. Assume these events are independent. What is the probability that Team A wins and Team B loses
The probability that Team A wins and Team B loses is 0.112.
Given that,
Suppose Team A has a 0.75 probability to win their next game and Team B has a 0.85 probability to win their next game.
Assume these events are independent.
We have to determine,
What is the probability that Team A wins and Team B loses?
According to the question,
In an independent event and probability, the outcomes in an experiment are termed as events. Ideally, there are multiple events like mutually exclusive events, independent events, dependent events, and more.
Team A has a 0.75 probability to win their next game,
And Team B has a 0.85 probability to win their next game.
Therefore,
The probability that Team A wins and Team B loses is
[tex]\rm The \ probability \ of \ A \ wins \ and \ team \ B \ loses = Probability \ of \ team \ A winning \ game \times (1- Probability \ of \ team B \ lose \ the \ game)\\\\ The \ probability \ of \ A \ wins \ and \ team \ B \ loses = 0.75 \times (1-0.85)\\\\ The \ probability \ of \ A \ wins \ and \ team \ B \ loses =0.75 \times 0.15\\\\ The \ probability \ of \ A \ wins \ and \ team \ B \ loses =0.112[/tex]
Hence, The probability that Team A wins and Team B loses is 0.112.
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Final answer:
The probability that Team A wins and Team B loses is calculated by multiplying the probability of Team A winning (0.75) with the probability of Team B losing (1 - 0.85). The result is 0.1125 or 11.25%.
Explanation:
To calculate the probability that Team A wins and Team B loses, we use the rules of independent events. The event of Team A winning has a probability of 0.75, and the event of Team B losing is the complement of Team B winning, which has a probability of 0.85. Since these are independent events, we multiply the probabilities:
P(Team A wins and Team B loses) = P(Team A wins) × P(Team B loses)
P(Team B loses) = 1 - P(Team B wins) = 1 - 0.85 = 0.15
Therefore, P(Team A wins and Team B loses) = 0.75 × 0.15 = 0.1125.
The probability that Team A wins and Team B loses is 0.1125, or 11.25%.
Recall that log 2 = 1 0 1 x+1 dx. Hence, by using a uniform(0,1) generator, approximate log 2. Obtain an error of estimation in terms of a large sample 95% confidence interval.
Answer:
∫101/(x+1)dx=(1−0)∫101/(x+1)dx/(1−0)=∫101/(x+1)f(x)dx=E(1/(x+1))
Where f(x)=1, 0
And then I calculated log 2 from the calculator and got 0.6931471806
From R, I got 0.6920717
So, from the weak law of large numbers, we can see that the sample mean is approaching the actual mean as n gets larger.
The probability of a train arriving on time and leaving on time is 0.8. The probability that the train arrives on time and leaves on time in 0.24. What is the probability that the train arrives on time, given that it leaves on time?
Answer:
0.9524
Step-by-step explanation:
Note enough information is given in this problem. I will do a similar problem like this. The problem is:
The Probability of a train arriving on time and leaving on time is 0.8.The probability of the same train arriving on time is 0.84. The probability of the same train leaving on time is 0.86.Given the train arrived on time, what is the probability it will leave on time?
Solution:
This is conditional probability.
Given:
Probability train arrive on time and leave on time = 0.8 Probability train arrive on time = 0.84 Probability train leave on time = 0.86Now, according to conditional probability formula, we can write:
[tex]P(Leave \ on \ time | arrive \ on \ time)[/tex] = P(arrive ∩ leave) / P(arrive)
Arrive ∩ leave means probability of arriving AND leaving on time, that is given as "0.8"
and
P(arrive) means probability arriving on time given as 0.84, so:
0.8/0.84 = 0.9524
This is the answer.
Beth looked through an old
cloth sack that contained small
plastic bags of yarn. There
were 8 bags of pink yarn, 11
white. 5 yellow, 3 green and 9
blue. What is the probability
that when Beth closes her eyes
and pulls out a bag, she will
get blue yarn? please help I have no clue how to do this and the next one is. what is the probability of her drawing white yarn will mark brainest ty please help godbless
Answer: blue is 25% chance
Step-by-step explanation:
8+11+5+3+9=36 then blue would be 9/36 of all the yarn and 9/36 converted into a percentage is 25%
Beth has a total of 36 yarn bags. The probability of her drawing a blue yarn is 9 out of 36, or 0.25. Similarly, the probability of her drawing a white yarn is 11 out of 36, or 0.306.
Explanation:The subject of this question is probability. The total number of yarn bags is the sum of all the different colors bags Beth has, which is 8 + 11 + 5 + 3 + 9 = 36 bags. The probability of an event occurring is determined by dividing the number of preferred events by the total number of outcomes. So, if Beth wants to pick a blue yarn, the number of favorable outcomes is 9 (blue yarn bags) and the total number of outcomes is 36 (total bags).
So, the probability of Beth picking a blue yarn is calculated as such 9/36 = 0.25.
Similarly, the probability of Beth picking a white yarn bag is calculated as 11/36 = 0.306. Hence, out of all the bags, she has a slightly higher chance of picking a white yarn bag.
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To prepare for surgery, Anne mixes an anesthetic solution using two different concentrations: 40 mL of 25% solution and 60 mL of 40% solution.
What is the concentration of the mixed solution?
Answer:
34%
Step-by-step explanation:
Amount of anesthetic in the 25% solution + amount of anesthetic in the 40% solution = amount of anesthetic in the mixed solution
0.25 (40) + 0.40 (60) = x (40 + 60)
10 + 24 = 100x
x = 0.34
The concentration of the mixed solution is 34%.
Compute the area of triangle, if x equals 4 more than 6. A) 10 B) 50 C) 100 D) 200 E) 400
Answer:
76cm²
Step-by-step explanation:
The triangle from all indication is an isosceles triangle:
Let x represent the side
∴ x = 4 - 6 = -2
x = -2
x + 2 = 0
Using the fomular, as area of triangle, that is:
Area = √s(s - a)(s - b)(s - c)
s = a + b+ c/2, where a = x + 2; b = x+ 2; c = x
∴ s = x + 2 + x + 2 + x/2 = 3x + 4/2
Area = √3x + 4/2( 3x + 4/2 - x - 2)(3x + 4/2 - x - 2)(3x + 4/2 - x)
= √3x + 4/2( x/2)(x/2)(x + 4/2)
= √3x + 4/2(x²/4)(x + 4)/2
Let's assume x = 12
∴ Area = √5760 = 76cm²
Answer:
Its C.100
Area of a triangle b*h/2 (6+4=10*2=20) 20*10=200/2 will give you 100.
Step-by-step explanation: