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11 months ago

What is the volume of a regular pyramid having a base area of 24 inches and height of 6 inches

Mathematics

2 answers:

Nimfa-mama [501]11 months ago

8 0

<h3>❁ Question -:</h3>

If the base area of a regular pyramid is 24 inches and the height is 6 inches. Find the volume of the regular pyramid ?

<h3>❁ Explanation -:</h3>

In this question we are provided with the base area that is 24 inches and it is also given that the height is 6 inches. We are asked to calculate the volume of the regular pyramid.

We know,

$✡ \: \small \underline{ \boxed{\sf {{Volume_{(pyramid)} = \dfrac{1}{3}×B × H}}}}$

Where,

B stand for base area.
H stand for height.

Substituting the values we get

$\small\frak{ Volume_{(pyramid)} = \dfrac{1}{3}×24 × 6}$

$\small\frak{ Volume_{(pyramid)} = 24 × 2}$

$\small\frak {Volume_{(pyramid)} =48 \: inches}$

$\small \underline{\boxed{ \frak{ Volume \: of \: a \: pyramid = 48 {inches }^{3} }}}$

Hence the volume of the pyramid is 48 inches ³.

NoTe : Always make sure that the volume will be in units³.

Anestetic [448]11 months ago

6 0

Answer:

V= 48 in^2

Step-by-step explanation:

Formula

Since the base has a known area, we do not need the full volume formula. This formula is

V = B * h / 3

B is the area of the base and h is the height measured from the top of the pyramid to the base meeting the base at right angles.

Givens

B = 24 in^2

h = 6 in

Solution

V = B * h / 3

V = 24 * 6/3

V= 48 in^2

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2 years ago

A high school principal wishes to estimate how well his students are doing in math. Using 40 randomly chosen tests, he finds tha

ollegr [7]

Answer:

99% confidence interval for the population proportion of passing test scores is [0.5986 , 0.9414].

Step-by-step explanation:

We are given that a high school principal wishes to estimate how well his students are doing in math.

Using 40 randomly chosen tests, he finds that 77% of them received a passing grade.

Firstly, the pivotal quantity for 99% confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of students received a passing grade = 77%

n = sample of tests = 40

p = population proportion

Here for constructing 99% confidence interval we have used One-sample z proportion test statistics.

So, 99% confidence interval for the population proportion, p is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99 {As the critical value of z at 0.5%

level of significance are -2.5758 & 2.5758}

P(-2.5758 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.5758) = 0.99

P( $-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

P( $\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

99% confidence interval for p = [ $\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.77-2.5758 \times {\sqrt{\frac{0.77(1-0.77)}{40} } }$ , $0.77+2.5758 \times {\sqrt{\frac{0.77(1-0.77)}{40} } }$ ]