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3 years ago

Help pls I need answers ASAP

Mathematics

1 answer:

pishuonlain [190]3 years ago

7 0

$\tt Step-by-step~explanation:$

To be able to solve this question, we need to know about the Pythagorean theorem: a² + b² = c². A and b represents the two sides that make the right angle, and C represents the hypotenuse, or the longest side, of the triangle.

$\tt Step~1:$

We know that the hypotenuse is the longest side of the triangle, so that means 9 is c. One side of the triangle is 6, and we have to solve for b. That means 6 is a. We can plug in our values to the formula like this:

$\tt a^2+b^2=c^2\\6^2+b+2=9^2\\36+b^2=81$

$\tt Step~2:$

Now that we have our equation down, we know what we have to do next: subtract 81 from 36 to get b².

$\tt 81-36=\sqrt{45}$

$\Large\boxed{\tt Our~final~answer:~b=\sqrt{45} ~(simplified:3\sqrt{5}) }$

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What is the area of the rectangle?<br><br> [1] un2

mezya [45]

(-2,3) to (2,-3)

6^2 + 4^2 = c^2

c = 7.21

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(-2,3) to (1,5)

2^2 + 3^2 = c^2

c = 3.61

————

3.61 * 7.21 = approx. 26.03

The area is 26.03un^2

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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.

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3 years ago

Read 2 more answers

In the arithmetic sequence {13, 6,-1,-8,...}, what is the common difference?

guapka [62]

<h3>Answer: -7</h3>

Explanation:

Pick any term. Subtract off the previous one to find the common difference.

term2 - term1 = 6-13 = -7
term3 - term2 = -1-6 = -7
term4 - term3 = -8-(-1) = -8+1 = -7

And so on. You only need to pick one of those to show as your steps to your teacher. However, doing all three subtractions is a good way to get practice in seeing how we have an arithmetic sequence. The common difference must be the same each time.

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2 years ago