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Katyanochek1 [597]
3 years ago
13

taneisha's family has signed up for a new cell phone plan.taneisha now has a limit on the number of texts she can send or receiv

e each month. she can text no more than 300 times in a month. what is the least number of texts she can make in a month
Mathematics
1 answer:
Snowcat [4.5K]3 years ago
5 0
0 What is this question lol
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On Monday, Raven ran 2 3 4 miles; Tuesday, 2 3 5 miles; Wednesday, 2 1 2 miles; Thursday, 2 5 8 miles; and Friday, 2 7 10 miles.
melisa1 [442]

Do you add them?? or is it multyplying 2 3 and 4??



5 0
3 years ago
Read 2 more answers
Help Please please please please !
il63 [147K]

Step-by-step explanation:

b.

The sum of two opposite interior angles equals one exterior angle

111°=60°+‹G

‹G=111°-60°=51°

c.

We use the same property above

3x+4x+5=68°

7x=68-5

7x=63

x=63/7=9

‹D=4(9)+5

‹D=41°

6 0
3 years ago
9x2+8x(5x-4) simplify
Svetach [21]

Answer:

40x --14

Step-by-step explanation:

Multiply 9 and 2 to get 18

Use distribute property

Multiply 8 by 5x -15

Subtract 32 from 18 = -14

4 0
2 years ago
Barron's reported that the average number of weeks an individual is unemployed is 18.5 weeks. Assume that for the population of
Ket [755]

Answer:

A) The sampling distribution for a sample size n=50 has a mean of 18.5 weeks and a standard deviation of 0.849.

B) P = 0.7616

C) P = 0.4441

Step-by-step explanation:

We assume that for the population of all unemployed individuals the population mean length of unemployment is 18.5 weeks and that the population standard deviation is 6 weeks.

A) We take a sample of size n=50.

The mean of the sampling distribution is equal to the population mean:

\mu_s=\mu=18.5

The standard deviation of the sampling distribution is:

\sigma_s=\dfrac{\sigma}{\sqrt{n}}=\dfrac{6}{\sqrt{50}}=0.849

B) We have to calculate the probability that the sampling distribution gives a value between one week from the mean. That is between 17.5 and 19.5 weeks.

We can calculate this with the z-scores:

z_1=\dfrac{X_1-\mu}{\sigma/\sqrt{n}}=\dfrac{17.5-18.5}{6/\sqrt{50}}=\dfrac{-1}{0.8485}=-1.179\\\\\\z_2=\dfrac{X_2-\mu}{\sigma/\sqrt{n}}=\dfrac{19.5-18.5}{6/\sqrt{50}}=\dfrac{1}{0.8485}=1.179

The probability it then:

P(|X_s-\mu_s|

C) For half a week (between 18 and 19 weeks), we recalculate the z-scores and the probabilities:

z=\dfrac{X-\mu}{\sigma/\sqrt{n}}=\dfrac{18-18.5}{6/\sqrt{50}}=\dfrac{-0.5}{0.8485}=-0.589

P(|X_s-\mu_s|

5 0
3 years ago
I just need help with these 2 questions !
Dahasolnce [82]

Answer:

-4/3

9/4

Step-by-step explanation:

hope this helped needed a longer answer lol

6 0
3 years ago
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