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Kobotan [32]
3 years ago
5

Find the slope of the line that passes through the points (-5,-6) and (5,10)

Mathematics
1 answer:
Montano1993 [528]3 years ago
8 0
Y=1.6x+2
that’s just what i got
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Solve the equation:<br> 5*4 / 10 + 4 =
salantis [7]

Answer:

6

Step-by-step explanation:

5 times 4 is 20

20 divided by 10 is 2

2 plus 4 is 6

7 0
3 years ago
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Whats this adding fraction answer
Lubov Fominskaja [6]
Its 5 for the first 1 1/3 for the second one
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3 years ago
The batting averages for seven baseball players are listed below. 0.176, 0.235, 0.272, 0.158, 0.201, 0.255, 0.303 What is the in
bonufazy [111]

Answer:

Interquartile Range= 0.079.

Step-by-step explanation:

Given,

The batting averages for seven baseball players are

0.176, 0.235, 0.272, 0.158, 0.201, 0.255, 0.303

First we have arrange the data in descending order or ascending order.

The ascending order of the data is

0.158, 0.176, 0.201, 0.235, 0.255, 0.272, 0.303

Median: The middle term of the data is the median of the data.

Here

Median = 0.235

First quartile: The median of the lower half of data.

The lower half of the data is 0.158, 0.176, 0.201

Q_1=0.176

Third quartile: The median of the upper half of data.

The upper half of the data is 0.201, 0.255, 0.303

Q_3=0.255

Interquartile: The difference between first quartile and third quartile.

Interquartile Range = Q_3-Q_1

                                  =(0.255- 0.176)

                                  =0.079

8 0
3 years ago
I need help on 5 a and b
coldgirl [10]
A:  63.75
B:  95.625 dollars
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3 years ago
According to the last census (2010), the mean number of people per household in the United States is LaTeX: \mu = 2.58 Assume a
Veseljchak [2.6K]

Answer:

P(2.50 < Xbar < 2.66) = 0.046

Step-by-step explanation:

We are given that Population Mean, \mu = 2.58 and Standard deviation, \sigma = 0.75

Also, a random sample (n) of 110 households is taken.

Let Xbar = sample mean household size

The z score probability distribution for sample mean is give by;

             Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

So, probability that the sample mean household size is between 2.50 and 2.66 people = P(2.50 < Xbar < 2.66)

P(2.50 < Xbar < 2.66) = P(Xbar < 2.66) - P(Xbar \leq 2.50)

P(Xbar < 2.66) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{2.66-2.78}{\frac{0.75}{\sqrt{110} } } ) = P(Z < -1.68) = 1 - P(Z  1.68)

                                                              = 1 - 0.95352 = 0.04648

P(Xbar \leq 2.50) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } \leq \frac{2.50-2.78}{\frac{0.75}{\sqrt{110} } }  ) = P(Z \leq  -3.92) = 1 - P(Z < 3.92)

                                                              = 1 - 0.99996 = 0.00004  

Therefore, P(2.50 < Xbar < 2.66) = 0.04648 - 0.00004 = 0.046

7 0
3 years ago
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