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3 years ago

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The volume of a pyramid varies jointly with the base area of the pyramid and its height. The volume of one pyramid is 35 cubic i

nches when its base area is 15 square inches and its height is 7 inches. What is the volume of a pyramid with a base area of 36 square inches and a height of 5 inches?

1 answer:

Alla [95]3 years ago

6 0

You first might want to divide 36 by 3, giving you 12. Multiply 12 by 5, which results in your answer of 60in^3. The equation for this is V = 1/3(blh)

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A new basketball is 40% off of its original price and is now selling for $36.00. What was the original price of the basketball b

VikaD [51]

Multiply 40% and 36.00 than add what you get to 36.

6 0

3 years ago

In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

Marina86 [1]

Answer:

(a) 0.94

(b) 0.20

(c) 90.53%

Step-by-step explanation:

From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let $p_1$ be the probability that a bearing meets the specification.

So, $p_1=0.9$

Sample size, $n_1=500$ , is large.

Let X represent the number of acceptable bearing.

Convert this to a normal distribution,

Mean: $\mu_1=n_1p_1=500\times0.9=450$

Variance: $\sigma_1^2=n_1p_1(1-p_1)=500\times0.9\times0.1=45$

$\Rightarrow \sigma_1 =\sqrt{45}=6.71$

(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.

So, $X\geq 440.$

Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{339.5-450}{6.71}=-1.56$ .

So, the probability that a given shipment is acceptable is

$P(z\geq-1.56)=\int_{-1.56}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}=0.94062$

Hence, the probability that a given shipment is acceptable is 0.94.

(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.

Denote the probability od acceptance of a shipment by $p_2$ .

$p_2=0.94$

The total number of shipment, i.e sample size, $n_2= 300$

Here, the sample size is sufficiently large to approximate it as a normal distribution, for which mean, $\mu_2$ , and variance, $\sigma_2^2$ .

Mean: $\mu_2=n_2p_2=300\times0.94=282$

Variance: $\sigma_2^2=n_2p_2(1-p_2)=300\times0.94(1-0.94)=16.92$

$\Rightarrow \sigma_2=\sqrt(16.92}=4.11$ .

In this case, X>285, so, by using the continuity correction, take x=285.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{285.5-282}{4.11}=0.85$ .

So, the probability that a given shipment is acceptable is

$P(z\geq0.85)=\int_{0.85}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}=0.1977$

Hence, the probability that a given shipment is acceptable is 0.20.

(c) For the acceptance of 99% shipment of in the total shipment of 300 (sample size).

The area right to the z-score=0.99

and the area left to the z-score is 1-0.99=0.001.

For this value, the value of z-score is -3.09 (from the z-score table)

Let, $\alpha$ be the required probability of acceptance of one shipment.

So,

$-3.09=\frac{285.5-300\alpha}{\sqrt{300 \alpha(1-\alpha)}}$

On solving

$\alpha= 0.977896$

Again, the probability of acceptance of one shipment, $\alpha$ , depends on the probability of meeting the thickness specification of one bearing.

For this case,

The area right to the z-score=0.97790

and the area left to the z-score is 1-0.97790=0.0221.

The value of z-score is -2.01 (from the z-score table)

Let $p$ be the probability that one bearing meets the specification. So

$-2.01=\frac{439.5-500 p}{\sqrt{500 p(1-p)}}$

On solving

$p=0.9053$

Hence, 90.53% of the bearings meet a thickness specification so that 99% of the shipments are acceptable.

8 0

3 years ago

Neptune’s average distance from the sun is 4.503 x 10e9 km. Mercury’s average distance from the sun is 5.791 x10e9 km.about how

Igoryamba

Answer:

77.76 times

Step-by-step explanation:

The average distance of Neptune from the sun

= 4.503 × 10 ⁹ k m .

and Mercury = 5.791 × 10 ⁷ k m .

Hence neptune is ( 4.503 × 10 ⁹) ÷ (5.791 × 10 ⁷ ) times farther from the sun than mercury

i.e.( $\frac{4.503}{5.791}$ ) × 10⁹⁻⁷ times

= 0.7776 × 10 ² times

= 77.76 times.

4 0

3 years ago

Simplify 4(х + 7) - 7х + 2. А. 11х + 9 В. 11х + 30 с. -3х + 9 D. -3х + 30

BartSMP [9]

Answer:

-3x+30

Step-by-step explanation:

$4\left(x+7\right)-7x+2$

Expand $4(x+7)= 4x+28$

$4x+28-7x+2$

Simplify $4x+28-7x+2$

$-3x+30$

7 0

3 years ago

Given the circle below, identify each choice that is a central angle.

Taya2010 [7]

can you upload a picture

3 0

2 years ago