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3 years ago

Identify 21, 46, and 57 as prime or composite. HELP!!

Mathematics

1 answer:

vlada-n [284]3 years ago

4 0

Answer:

Composite, Composite, composite

Step-by-step explanation:

all of them can be divided

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4+5(p-1) how to expand brackets and then simplify

Valentin [98]

Step-by-step explanation:

these are multiplications.

you could also write

4 + 5×(p - 1)

now, we need to calculate the contents of brackets (if we can), then do the multiplications and divisions, before we can do the then remaining additions and subtractions.

if we cannot do a calculation directly (because there is a variable involved), we need to do and document the single steps for the individual parts involved.

so,

4 + 5×(p - 1) = 4 + 5×p + 5×-1 = 4 + 5p - 5 = 5p - 1

remember, a multiplication of 2 expressions is done by multiplying every term of one expression with every term of the other expression and adding the results up (by considering their individual signs, of course).

4 0

2 years ago

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 $\bar x_{critical}$