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2 years ago

Factor -28x +20x show work

1 answer:

8 0

Let's simplify step-by-step.−28x+20x

ombine Like Terms:=−28x+20x=(−28x+20x)=−8x

Answer:=−8x

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garri49 [273]

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2 years ago

The height of a 12 year old boy increased by 17% last year and by 12% this year. What is the total increase in percent terms?

OverLord2011 [107]

He increased by 29% because you just add up the percentage

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2 years ago

Read 2 more answers

Solve the equation. x6−7=5 The first step is to___ on both sides of the equation. The second step is to___ on both sides of the

gogolik [260]

Answer:

x=2

Step-by-step explanation:

x6−7=5

x6=7+5

x6=12|:6

x=2

3 0

2 years ago

A food company sells salmon to various customers. The mean weight of the salmon is 44 lb with a standard deviation of 3 lbs. The

TiliK225 [7]

Correct question:

A food company sells salmon to various customers. The mean weight of the salmon is 44 lb with a standard deviation of 3 lbs. The company ships them to restaurants in boxes of 9 salmon, to grocery stores in cartons of 16 salmon, and to discount outlet stores in pallets of 64 salmon. To forecast costs, the shipping department needs to estimate the standard deviation of the mean weight of the salmon in each type of shipment. Complete parts (a) and (b) below.

a. Find the standard deviations of the mean weight of the salmon in each type of shipment.

b. The distribution of the salmon weights turns out to be skewed to the high end. Would the distribution of shipping weights be better characterized by a Normal model for the boxes or pallets?

Answer:

Given:

Mean, u = 44

Sd = 3

The company ships in boxes of 9, cartons of 16 and pallets of 64.

a) For the standard deviations of the mean weight of the salmon in each type of shipment, lets use the formula: $\frac{s.d}{\sqrt{u}}$

i) For the standard deviation of the mean weight of salmon in boxes of 9, we have:

$\frac{s.d}{\sqrt{u}}$

$= \frac{3}{\sqrt{9}}$

$= \frac{3}{3} = 1$

The standard deviation = 1

ii) For the standard deviation of the mean weight of salmon in cartons of 16, we have:

$\frac{s.d}{\sqrt{u}}$

$= \frac{3}{\sqrt{16}}$

$= \frac{3}{4} = 0.75$

Standard deviation = 0.75

iii) For the standard deviation of the mean weight of salmon in pellets of 64, we have:

$\frac{s.d}{\sqrt{u}}$

$= \frac{3}{\sqrt{64}}$

$= \frac{3}{8} = 0.375$

Standard deviation = 0.375

b) The distribution of shipping weights would be better characterized by a Normal model for the pallets, because regardless of the underlying distribution, the sampling distribution of the mean approaches the Normal model as the sample increases.

5 0

2 years ago

What is the answer to this?

NARA [144]

Answer:

See work for 3 answers

Step-by-step explanation:

You can add the exponents

$(-8)^{12}$