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2 years ago

15

The support pole of the tent shown forms one leg of a right triangle. One side of the tent forms the hypotenuse of the right tri

angle. Find the length of the base of the tent.
Someone please helpppp!!!!1

1 answer:

svetoff [14.1K]2 years ago

8 0

Ans=24
You need to use Pythagorean theorem.
A^2+b^2=c^2
X^2+32^2=40^2
X^2+1024=1600
1600-1024=576
576=x^2
Square root of 576=x=24

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rjkz [21]

Answer:

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Step-by-step explanation:

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2 years ago

I need help on this 2 problems please help

Digiron [165]

Answer:

3. 294 m²

4. 185,856 mm²

Step-by-step explanation:

To find the surface area of any figure, you can simply find the sum of the areas of all the sides. In question 3, the figure is a triangular prism that has two triangular bases and three rectangular sides. The measurements for the triangles are b = 9m and h = 6m. The formula for the area of a triangle is base times height divided by 2, or 9 x 6 = 54/2 = 27 m². However, since there are two triangular bases, the area for both is 54 m². The measurements for the other three rectangles are given:

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6 0

3 years ago

Marcos purchased a sail boat for $56,900. He anticipates each year the boat will decrease in value by 8.1% from the previous yea

Varvara68 [4.7K]

Answer: $56,900\left( 0.919\right)^x$

Step-by-step explanation:

Given

Marcos purchased a sailboat for $\$56,900$

Each year the boat will decrease in value by $8.1\%$

After 1 year it is

$\Rightarrow 56,900-56,900\times 0.081\\\Rightarrow 56,900\left( 1-0.081\right)\\$

after another year it becomes

$\Rightarrow 56,900\left( 1-0.081\right)-56,900\left( 1-0.081\right)\times 0.081\\\Rightarrow 56,900\left( 1-0.081\right)\cdot \left( 1-0.081\right)\\\Rightarrow 56,900\left( 1-0.081\right)^2$

After $x$ years it is

$\Rightarrow 56,900\left( 1-0.081\right)^x$

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2 years ago

Suppose in the population, the Anger-Out score for men is two points higher than it is for women. The population variances for m

tatuchka [14]

Answer:

a)= 2

b) 6.324

c) P= 0.1217

Step-by-step explanation:

a) The mean of the sampling distribution of X`1- X`2 denoted by ux`-x` = u1-u2 is equal to the difference between population means i.e = 2 ( given in the question)

b) The standard deviation of the sampling distribution of X`1- X`2 ( standard error of X`1- X`2) denoted by σ_X`1- X`2 is given by

σ_X`1- X`2 = √σ²/n1 +σ²/n2

Var ( X`1- X`2) = Var X`1 + Var X`2 = σ²/n1 +σ²/n2

so

σ_X`1- X`2 =√20 +20 = 6.324

if the populations are normal the sampling distribution X`1- X`2 , regardless of sample sizes , will be normal with mean u1-u2 and variance σ²/n1 +σ²/n2.

Where as Z is normally distributed with mean zero and unit variance.

If we take X`1- X`2= 0 and u1-u2= 2 and standard deviation of the sampling distribution = 6.324 then

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P(-0.31625<z<0)= 0.1217

The probability would be 0.1217

8 0

2 years ago

The following prices, in dollars, of 7.5-cubic-foot refrigerators were recorded from a random sample. 314 305 344 283 285 310 3

Mariana [72]

Answer:

$t=\frac{310.9-300}{\frac{31.09}{\sqrt{10}}}=1.109$

The degrees of freedom are given by:

$df=n-1=10-1=9$

And the p value would be given by:

$p_v =P(t_{9}>1.109)=0.148$

Since the p value is higher than the significance level of 0.05 we don't have enough evidence to conclude that the true mean is significantly higher than $300 because we FAIL to reject the null hypothesis.

Step-by-step explanation:

Information given

314 305 344 283 285 310 383 285 300 300

We can calculate the sample mean and deviation with the following formula:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

$s =\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}$

$\bar X=310.9$ represent the sample mean

$s=31.09$ represent the standard deviation for the sample

$n=10$ sample size

$\mu_o =300$ represent the value to test

$\alpha=0.05$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to verify if the true mean is greater than 300, the system of hypothesis would be:

Null hypothesis: $\mu \leq 300$

Alternative hypothesis: $\mu > 300$

The statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info given we got:

$t=\frac{310.9-300}{\frac{31.09}{\sqrt{10}}}=1.109$

The degrees of freedom are given by:

$df=n-1=10-1=9$

And the p value would be given by:

$p_v =P(t_{9}>1.109)=0.148$

Since the p value is higher than the significance level of 0.05 we don't have enough evidence to conclude that the true mean is significantly higher than $300 because we FAIL to reject the null hypothesis.

6 0

2 years ago