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3 years ago

Match each whole number with a rational, exponential expression.

Mathematics

1 answer:

Katyanochek1 [597]3 years ago

4 0

$1.\\9=3^2\\2.\\196=14^2\\3.\\256=16^2=4^4=2^8\\4.\\225=15^2\\5.\\16=4^2=2^4\\6.\\25=5^2$

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Which is the value of the expresson?

Katyanochek1 [597]

The solution to your problem is 8 (:

3 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

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

Morgan is reupholstering an old chair with two circular cushions. Each cushion has a circumference of 7. 85 feet. He will need t

kramer

The total fabric Morgan needs to cover the cushions will be 19.64 square feet.

It is given that each cushion has a circumference of 7.85 feet.

It is known that the circumference of a circle = 2πr, where 'r' is the radius of the circle.

2πr = 7.85

r = 1.25

<h3>What is an area of a circle?</h3>

area of a circle is πr².

now, we can calculate the area = π× 1.25² = 4.91 square feet

since each circular cushion has two circular faces so the total number of faces Morgan needs to cover will be 4.

Therefore, the total fabric needed= 4×4.91 = 19.64 square feet.

to get more about the area of a circle, refer to the link:

brainly.com/question/12069107

8 0

2 years ago

Look at the picture ASAP

goldenfox [79]

Using the Laws of Cosine, we obtain $C^2 = 55^2+108^2-2(108)(55)*cos(59)$ , which c approximately equals 153.44. So Line AB ≈ 153.44 meters.

8 0

3 years ago

Find the slope between these two points: (3, -10) and (-2, -30). Show all of your work. what the answer

Harlamova29_29 [7]

$\huge\text{$m=\boxed{4}$}$

Hey there! Start with the slope formula, where $(x_1,y_1)$ and $(x_2,y_2)$ are the two known points.

$\begin{aligned}m&=\dfrac{y_2-y_1}{x_2-x_1}\\&=\frac{-30-(-10)}{-2-3}\end{aligned}$

Simplify.

$\begin{array}{c|l}\textbf{Solving}&\textbf{Reason}\\\cline{1-2}\\m=\dfrac{-30+10}{-2-3}&x-(-y)=x+y\\\\m=\dfrac{-20}{-5}&\text{Addition and subtraction}\\\\m=\dfrac{20}{5}&\text{The negatives cancel out}\\\\m=\dfrac{4}{1}&\text{Divide the numerator and denominator by $5$}\\\\m=\boxed{4}&\dfrac{x}{1}=x\end{aligned}$

3 0

3 years ago

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