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3 years ago

I have never learned this before please help

Mathematics

1 answer:

jek_recluse [69]3 years ago

6 0

Answer:

V =165 ft^3

Step-by-step explanation:

V = Bh Where B is the area of the base and h is the height

B is the area of the triangular base

B = 1/2 (5) *6

B = 15 ft^2

V = 15 * 11

V =165 ft^3

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3 years ago

NO LINKS!! Explain your answer (show the support by showing the changes in x and y on your table). If the relationship is linear

ivolga24 [154]

Answer:

relationship: inverse
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Step-by-step explanation:

You are correct that the relationship is inverse, and that the product of x and y is a constant, 25.

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2 years ago

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this is a blueprint drawing of the stage area at Millennium Park the length of one square on the grid is equal to 5 feet accordi

dmitriy555 [2]

Answer:

No, diagonal of square C is 70.71ft

Step-by-step explanation:

Given in the question that length of one square = 5 ft

Length of Square C = 10 small squares

1 square = 5 ft

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As we know that in a square all side are of equal length and make an angle of 90° at each corner.

So, by using pythagorus theorem

Diagonal of square² = side² + side²

Diagonal of square² = 2side²

Diagonal of square = √(2side²)

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3 years ago

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zmey [24]

Answer:

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2 years ago

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We wish to give a 90% confidence interval for the mean value of a normally distributed random variable. We obtain a simple rando

coldgirl [10]

Answer: $(9.27025,\ 11.12975)$

Step-by-step explanation:

Given : Sample size : n= 9

Degree of freedom = df =n-1 =8

Sample mean : $\overline{x}=10.2$

sample standard deviation : $s= 1.5$

Significance level ; $\alpha= 1-0.90=0.10$

Since population standard deviation is not given , so we use t- test.

Using t-distribution table , we have

Critical value = $t_{\alpha/2, df}=t_{0.05 , 8}=1.8595$

Confidence interval for the population mean :

$\overline{x}\pm t_{\alpha/2, df}\dfrac{s}{\sqrt{n}}$

90% confidence interval for the mean value will be :

$10.2\pm (1.8595)\dfrac{1.5}{\sqrt{9}}$

$10.2\pm (1.8595)\dfrac{1.5}{3}$

$10.2\pm (1.8595)(0.5)$

$10.2\pm (0.92975)$

$(10.2-0.92975,\ 10.2+0.92975)$

$(9.27025,\ 11.12975)$

Hence, the 90% confidence interval for the mean value= $(9.27025,\ 11.12975)$

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3 years ago