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2 years ago

Choose the amounts of sales that would give Shonda's business a positive

Mathematics

2 answers:

Hitman42 [59]2 years ago

8 0

Answer:

Hello! Your answer would be, BELOW!

Step-by-step explanation:

$26

$21

$29

$17

Hope I helped! Ask me anything if you have more questions! Have a nice mourning! Brainiest plz! Hope you make an 100%! -Amelia♥

kicyunya [14]2 years ago

8 0

Hi! $26,$21,$29,$17 would work.

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-3 1/3b =5 what is the value of b?

zysi [14]

Answer:

b = (-3)/2

Step-by-step explanation:

Solve for b:

(-10 b)/3 = 5

Multiply both sides of (-10 b)/3 = 5 by -3/10:

(-3 (-10) b)/(10×3) = (-3)/10×5

(-3)/10×(-10)/3 = (-3 (-10))/(10×3):

(-3 (-10))/(10×3) b = (-3)/10×5

(-3)/10×5 = (-3×5)/10:

(-3 (-10) b)/(10×3) = (-3×5)/10

(-10)/10 = (10 (-1))/10 = -1:

(-3-1 b)/3 = (-3×5)/10

(-3)/3 = (3 (-1))/3 = -1:

--1 b = (-3×5)/10

(-1)^2 = 1:

b = (-3×5)/10

5/10 = 5/(5×2) = 1/2:

Answer: b = (-3)/2

5 0

3 years ago

Read 2 more answers

A young oak tree is planted when it is 5 cm tall. It grows 3 cm every month. Write an equation to describe its growth

Naily [24]

Answer:

y=3x+5

Since it's already 5 cm we have +5

every month it would go up by 3 cm so we have 3 and x is to represent the # of months.

8 0

2 years ago

8.23 A manufacturing company produces electric insulators. You define the variable of interest as the strength of the insulators

Kitty [74]

Answer:

The 95% confidence interval estimate for the population mean force is (1691, 1755).

Step-by-step explanation:

According to the Central Limit Theorem if we have an unknown population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample mean will be approximately normally.

The sample selected here is <em>n</em> = 30.

Thus, the sampling distribution of the sample mean will be normal.

Compute the sample mean and standard deviation as follows:

$\bar x=\frac{1}{n}\sum x=\frac{1}{30}\times 51702=1723.4\\\\s=\sqrt{\frac{1}{n-1}\sum (x-\bar x)^{2}}=\sqrt{\frac{1}{30-1}\times 232561.2}=89.55$