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3 years ago

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A triangular pyramid with a height of 9 inches has a volume of 63 cubic inches. If the height of the triangular base is 6 inches

,
what is the base length of the triangular base? (Recall the formula V - 3 Bh)
A) 2 2/3 in
B) 7 in
C) 10 1/2 in
D) 21 in

1 answer:

Bond [772]3 years ago

3 0

Option D: 21 in is the base length of the triangular base.

Explanation:

Given that a triangular pyramid with a height of 9 inches has a volume of 63 cubic inches.

The height of the triangular base is 6 inches.

We need to determine the base length of the triangular pyramid.

The base length of the triangular pyramid can be determined using the formula,

$Volume =\frac{1}{3} \times Bh$

Substituting $Volume=63$ and $Height=9$ in the above formula, we get,

$63 =\frac{1}{3} \times B(9)$

Simplifying the terms, we get,

$63 =3B$

Dividing both sides by 3, we have,

$21=B$

Thus, the base length of the triangular pyramid is 21 in

Hence, Option D is the correct answer.

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How many times must we toss a coin to ensure that a 0.95-confidence interval for the probability of heads on a single toss has l

musickatia [10]

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(1) 97

(2) 385

(3) 9604

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The (1 - <em>α</em>) % confidence interval for population proportion is:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

The margin of error in this interval is:

$MOE= z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

The formula to compute the sample size is:

$\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}$

(1)

Given:

$\hat p = 0.50\\MOE=0.1\\z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use the <em>z</em>-table for the critical value.

Compute the value of <em>n</em> as follows:

$\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}\\=\frac{1.96^{2}\times0.50\times(1-0.50)}{0.1^{2}}\\=96.04\\\approx97$

Thus, the minimum sample size required is 97.

(2)

Given:

$\hat p = 0.50\\MOE=0.05\\z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use the <em>z</em>-table for the critical value.

Compute the value of <em>n</em> as follows:

$\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}\\=\frac{1.96^{2}\times0.50\times(1-0.50)}{0.05^{2}}\\=384.16\\\approx385$

Thus, the minimum sample size required is 385.

(3)

Given:

$\hat p = 0.50\\MOE=0.01\\z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use the <em>z</em>-table for the critical value.

Compute the value of <em>n</em> as follows:

$\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}\\=\frac{1.96^{2}\times0.50\times(1-0.50)}{0.01^{2}}\\=9604$

Thus, the minimum sample size required is 9604.

8 0

3 years ago

riadik2000 [5.3K]

Answers:

Domain is $0 \le d \le 100$

Range is $0 \le F(d) \le 150$

==============================================

Explanation:

The most money made from selling drinks is $150. Divide this over the cost per drink to get 150/(1.50) = 100; indicating that at most 100 drinks are sold per day. This is the largest that d can get because d represents the number of drinks sold.

The smallest d can get is d = 0 to mean that no drinks are sold.

In short: d is between 0 and 100, including both endpoints. We write $0 \le d \le 100$ to indicate this. This is the domain because the domain represents all the possible inputs allowed.

---------------------------------------------------

The range is the set of allowed outputs.

If we plugged in d = 0, then you would find F(d) = 0 as well. If you don't sell any drinks, then you earn $0. This is the smallest item in the range.

On the other side of things, the largest item in the range is 150 because this value was given to us. It's the upper limit or ceiling value of how much money is made from drinks. You can also find this by plugging d = 100, the largest domain value, into the function to get F(d) = 150.

Therefore the range is $0 \le F(d) \le 150$ to indicate that F(d) is between 0 and 150 inclusive of both endpoints.

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3 years ago