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3 years ago

The sum of 2 numbers is 52. Express the product of the numbers, P, as a function of one of the numbers, x.

Mathematics

1 answer:

kolezko [41]3 years ago

8 0

Answer:

The sum of 2 numbers is 52. Express the product of the numbers, P, as a function of one of the numbers, x.

the answer is p(x) = -x² + 52x

Step-by-step explanation:

let the two numbers be x and y

the sum of the two numbers = x + y

≡ x + y = 52

make y the subject of equation

y= 52 - x

to express the product of the numbers, P, as a function of one of the numbers, x. implies that

P(x) = x(-x + 52)

finally,

p(x) = -x² + 52x

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2 years ago

Which situation can be represented by the inequality 4x − 25 < 125?

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3 years ago

What is the slope of y-5 = 4 ( x+3)

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Step-by-step explanation:

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3 years ago

Read 2 more answers

A consumer products company is formulating a new shampoo and is interested in foam height (in mm). Foam height is approximately

Genrish500 [490]

Answer:

a) 0.057

b) 0.5234

c) 0.4766

Step-by-step explanation:

To find the p-value if the sample average is 185, we first compute the z-score associated to this value, we use the formula

$z=\frac{\bar x-\mu}{\sigma/\sqrt N}$

where

$\bar x=mean\; of\;the \;sample$

$\mu=mean\; established\; in\; H_0$

$\sigma=standard \; deviation$

N = size of the sample.

So,

$z=\frac{185-175}{20/\sqrt {10}}=1.5811$

$\boxed {z=1.5811}$

As the sample suggests that the real mean could be greater than the established in the null hypothesis, then we are interested in the area under the normal curve to the right of 1.5811 and this would be your p-value.

We compute the area of the normal curve for values to the right of 1.5811 either with a table or with a computer and find that this area is equal to 0.0569 = 0.057 rounded to 3 decimals.

So the p-value is

$\boxed {p=0.057}$

Since the z-score associated to an α value of 0.05 is 1.64 and the z-score of the alternative hypothesis is 1.5811 which is less than 1.64 (z critical), we cannot reject the null, so we are making a Type II error since 175 is not the true mean.

We can compute the probability of such an error following the next steps:

Step 1

Compute $\bar x_{critical}$

$1.64=z_{critical}=\frac{\bar x_{critical}-\mu_0}{\sigma/\sqrt{n}}$

$\frac{\bar x_{critical}-\mu_0}{\sigma/\sqrt{n}}=\frac{\bar x_{critical}-175}{6.3245}=1.64\Rightarrow \bar x_{critical}=185.3721$

So we would make a Type II error if our sample mean is less than 185.3721.

Step 2

Compute the probability that your sample mean is less than 185.3711

$P(\bar x < 185.3711)=P(z< \frac{185.3711-185}{6.3245})=P(z$

So, the probability of making a Type II error is 0.5234 = 52.34%

The power of a hypothesis test is 1 minus the probability of a Type II error. So, the power of the test is

1 - 0.5234 = 0.4766

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3 years ago

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Lerok [7]

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3 years ago