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2 years ago

11

. Printer A can print 1/2 page per second. Printer B can print 1/3 page per second. Jacob has 5 pages to print. How long would i

t take printer A to print 5 pages

1 answer:

Paul [167]2 years ago

6 0

10 seconds.

2 seconds for one page.

2 times 5=10

The answer=10!!!!!!!!!!!!!!

I hope that helps!:)

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Hong Tang's gross weekly salary is $615. His weekly federal withholding is $55.

Ulleksa [173]

too much math for my mental health

7 0

2 years ago

What is the product of a+3 and −2a2+15a+6b2?

rewona [7]

For this case we must find the product of the following expression:

$(a + 3) (- 2a ^ 2 + 15a + 6b ^ 2)$

We apply distributive property term to term:

$(a + 3) (- 2a ^ 2 + 15a + 6b ^ 2) =\\-2a ^ 2 * a + 15a * a + 6b ^ 2 * a-6a ^ 2 + 45a + 18b ^ 2 =\\-2a ^ 3 + 15a ^ 2 + 6ab ^ 2-6a ^ 2 + 45a + 18b ^ 2=$

We add similar terms:

$-2a ^ 3 + (15-6) a ^ 2 + 6ab ^ 2 + 45a + 18b ^ 2 =\\-2a ^ 3 + 9a ^ 2 + 6ab ^ 2 + 45a + 18b ^ 2 =$

Finally, the expression is as:

$-2a ^ 3 + 9a ^ 2 + 6ab ^ 2 + 45a + 18b ^ 2$

Answeer:

$-2a ^ 3 + 9a ^ 2 + 6ab ^ 2 + 45a + 18b ^ 2$

5 0

2 years ago

Use the graph to write an explicit function to represent the data and determine how much money Amy earned in week 5.

zalisa [80]

Answer:

3rd option

$f(n)=5(3)^{n-1}$

$f(5)=405$

Step-by-step explanation:

So we are given the following points:

(1,5)

(2,15)

(3,45)

(4,135)

This is a geometric sequence because there is a common ratio, 3. That is you can keep multiply 3 to a previous y-coordinate to get the next y-coordinate.

The formula for a geometric sequence is $f(n)=a_1 \cdot r^{n-1}$

where $a_1$ is the first term and r is the common ration.

So we have

$f(n)=5 \cdot (3)^{n-1}$ .

If you want to know the fifth term, just plug in 5:

$f(5)=5 \cdot (3)^{5-1}$

Simplifying:

$f(5)=5 \cdot 3^4$

$f(5)=5 \cdot 81$

$f(5)=405$

3 0

2 years ago

Read 2 more answers

1) The roots of the equation 9x2 + 3x – 4 = 0 are

Wittaler [7]

A !! hope this helps

7 0

2 years ago

Read 2 more answers

The XO Group Inc. conducted a survey of 13,000 brides and grooms married in the United States and found that the average cost of

slavikrds [6]

Answer:

a) 0.0392

b) 0.4688

c) At least $39,070 to be among the 5% most expensive.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 29858, \sigma = 5600$

a. What is the probability that a wedding costs less than $20,000 (to 4 decimals)?

This is the pvalue of Z when X = 20000. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{20000 - 29858}{5600}$

$Z = -1.76$

$Z = -1.76$ has a pvalue of 0.0392.

So this probability is 0.0392.

b. What is the probability that a wedding costs between $20,000 and $30,000 (to 4 decimals)?

This is the pvalue of Z when X = 30000 subtracted by the pvalue of Z when X = 20000.

X = 30000

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{30000 - 29858}{5600}$

$Z = 0.02$

$Z = 0.02$ has a pvalue of 0.5080.

X = 20000

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{20000 - 29858}{5600}$

$Z = -1.76$

$Z = -1.76$ has a pvalue of 0.0392.

So this probability is 0.5080 - 0.0392 = 0.4688

c. For a wedding to be among the 5% most expensive, how much would it have to cost (to the nearest whole number)?

This is the value of X when Z has a pvalue of 0.95. So this is X when Z = 1.645.

$Z = \frac{X - \mu}{\sigma}$

$1.645 = \frac{X - 29858}{5600}$

$X - 29858 = 5600*1.645$

$X = 39070$

The wedding would have to cost at least $39,070 to be among the 5% most expensive.

5 0

2 years ago