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Fiesta28 [93]
3 years ago
9

The highest scorer of the​ women's basketball championship was Jessica Bradley. She scored 157 more points than Tina​ Harner, he

r teammate.​ Together, Bradley and Harner scored 1623 points. How many points did each player score during the​ championship?
Mathematics
1 answer:
iren2701 [21]3 years ago
3 0

Answer:

It is not possible to score that many points. Only Wilt Chamberlain can reach 100 points. lol

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Shirley wants to buy a skateboard for $64. She has $98 in her account. She spent $10.85 to buy stationary. She also wants to buy
klemol [59]

she had 98 but spent 10.85 so she now has:

98-10.85 = 87.15

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87.15 -64 = 23.15 is what she can spend on cookies

23.15 / 1.65 = 14.03 she can't buy more than 14 cookies

 so answer is c. n ≤ 14

5 0
4 years ago
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Using subtraction what is 4 1/6-1 5/6
krek1111 [17]

Answer:

4

Step-by-step explanation:

the answer is 4 because 2.33333333333

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4 0
3 years ago
Find the median and mean of the data set below:<br> 47,15, 6, 49, 45, 30
Georgia [21]

Answer:

The median of the data set:

6, 15, 30, 45, 47, 49

The middle numbers are 30 and 45 so

You have to do 30 + 45

= 75

Then do 75 ÷ 2

= 37.5

The mean of the data set:

47, 15, 6, 49, 45, 30 (add them all)

= 192

Then divide 192 by 6 (because there are 6 numbers)

192 ÷ 6

= 32

So the median is 37.5 and the mean is 32.

Step-by-step explanation:

Hope this helps!

From your neighborhood softie :)

5 0
2 years ago
In the library on a university campus, there is a sign in the elevator that indicates a limit of 16 persons. Furthermore, there
Vedmedyk [2.9K]

Answer:

a) 151lb.

b) 6.25 lb

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

In this problem, we have that:

\mu = 151, \sigma = 25, n = 16

So

a) The expected value of the sample mean of the weights is 151 lb.

(b) What is the standard deviation of the sampling distribution of the sample mean weight?

This is s = \frac{\sigma}{\sqrt{n}} = \frac{25}{\sqrt{16}} = 6.25

8 0
3 years ago
Look at the table showing different prices for oranges. 
MrMuchimi
2.25 for 6 would be 0.375 per one of them
1.30 for 4 would be 0.325 per one of them
0.97 for 2 would be 0.485 per one of them
5 0
3 years ago
Read 2 more answers
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