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2 years ago

6

A school is printing tshirts for their students. They have received 2 pricesfrom 2 companies. Company A says they will print 16

tshirts for $40. Company B days they will print 84 tshirts for $140. Which company is selling t shirts at a better price?

1 answer:

Goshia [24]2 years ago

8 0

Company B is selling at a better price.

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What is the midpoint segment with endpoints j(-3,-2),K(4,-6)?

olganol [36]

Idk sry I wish I did know but I don’t

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3 years ago

Using the two functions listed below, insert numbers in place of the letters a, b, c, and d so that f(x) and g(x) are inverses.

blondinia [14]

I hope this helps you

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3 years ago

If the sum of four consecutive numbers is 854, what is the third one?

lukranit [14]

Answer:

214

Step-by-step explanation:

The set up would be x + (x + 1) + (x + 2) + (x + 3) = 854, naturally as the numbers are consecutive. Solving for x:

x + (x + 1) + (x + 2) + (x + 3) = 854,

x + x + 1 + x + 2 + x + 3 = 854,

4x + 1 + 2 + 3 = 854,

4x + 6 = 854,

4x = 854 - 6 = 848,

x = 848/4 = 212

The third number then should be 212 + 2 = 214

7 0

3 years ago

Andre wants to save $40 to buy a gift for his dad. Andre’s neighbor will pay him weekly to mow the lawn, but Andre always gives

Dmitry_Shevchenko [17]

(I could answer more confidently if I had more context to the question)

I believe the answer to the question is Andre's neighbor pays Andre 10 dollars a week to mow his lawn.

Hope this helps, if not, comment below please!!

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2 years ago

Use the fact that the mean of a geometric distribution is μ= 1 p and the variance is σ2= q p2. A daily number lottery chooses th

butalik [34]

Answer:

a). The mean = 1000

The variance = 999,000

The standard deviation = 999.4999

b). 1000 times , loss

Step-by-step explanation:

The mean of geometric distribution is given as , $\mu = \frac{1}{p}$

And the variance is given by, $\sigma ^2=\frac{q}{p^2}$

Given : $p=\frac{1}{1000}$

= 0.001

The formulae of mean and variance are :

$\mu = \frac{1}{p}$

$\sigma ^2=\frac{q}{p^2}$

$\sigma ^2=\frac{1-p}{p^2}$

a). Mean = $\mu = \frac{1}{p}$

= $\mu = \frac{1}{0.001}$

= 1000

Variance = $\sigma ^2=\frac{1-p}{p^2}$

= $\sigma ^2=\frac{1-0.001}{0.001^2}$

= 999,000

The standard deviation is determined by the root of the variance.

$\sigma = \sqrt{\sigma^2}$

= $\sqrt{999,000}$ = 999.4999

b). We expect to have play lottery 1000 times to win, because the mean in part (a) is 1000.

When we win the profit is 500 - 1 = 499

When we lose, the profit is -1

Expected value of the mean μ is the summation of a product of each of the possibility x with the probability P(x).

$\mu=\Sigma\ x\ P(x)= 499 \times 0.001+(-1) \times (1-0.001)$

= $ 0.50

Since the answer is negative, we are expected to make a loss.

4 0

2 years ago