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Alex787 [66]
3 years ago
11

In the drawer you have 8 black socks and 5 yellow socks, if you pick 2 socks what is the probability that pick out a black sock

and then a yellow sock , you don’t need to replace the first sock
Mathematics
1 answer:
Nataliya [291]3 years ago
7 0

Answer:

10/39

Step-by-step explanation:

The chances of picking a black sock on first try are 8/13, and the chances of picking out a yellow sock without replacing the black sock are 5/12. If you multiply 8/13 and 5/12, you get 40/156, and if you simplify, you will get 10/39 as your answer.

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Simplify the linear expression. −2x+3(−5x+1) Enter your answer in the box.
valina [46]

Answer:

3/17

Step-by-step explanation:

-2x + 3(-5x + 1)

-2x -15x +3

-17x + 3

17x = 3

x = 3/17

6 0
3 years ago
Divide 24xy²z³ by 6yz²​
liberstina [14]

Answer: 4xyz

Step-by-step explanation:

24xy²z³ / 6yz² =4xy²⁻¹z³⁻² = 4xyz

3 0
3 years ago
Rent and other associated housing costs, such as utilities, are an important part of the estimated costs of attendance at colleg
zhenek [66]

Answer:

96% confidence interval estimate for the mean monthly rent of all unmarried BYU students in winter 2018 is  [335.89 , 356.10].

Step-by-step explanation:

We are given that a group of researchers at the BYU Off-Campus Housing department want to estimate the mean monthly rent that unmarried BYU students paid during winter 2018.

During March 2018, they randomly sampled 314 BYU students and found that on average, students paid $346 for rent with a standard deviation of $86.

So, the pivotal quantity for 95% confidence interval for the average age is given by;

           P.Q. = \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where, \bar X = average rent paid by 314 BYU students = $346

            s = sample standard deviation = $86

            n = sample of students = 314

            \mu = population mean monthly rent of all unmarried BYU students

So, 96% confidence interval for the population mean monthly rent, \mu is ;

P(-2.082 < t_3_1_3 < 2.082) = 0.96

P(-2.082 < \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } < 2.082) = 0.96

P( -2.082 \times {\frac{s}{\sqrt{n} } } < {\bar X - \mu} < 2.082 \times {\frac{s}{\sqrt{n} } } ) = 0.96

P( \bar X -2.082 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X +2.082 \times {\frac{s}{\sqrt{n} } } ) = 0.96

96% confidence interval for \mu = [ \bar X -2.082 \times {\frac{s}{\sqrt{n} } } , \bar X +2.082 \times {\frac{s}{\sqrt{n} } } ]

                                                = [ 346 -2.082 \times {\frac{86}{\sqrt{314} } } , 346 +2.082 \times {\frac{86}{\sqrt{314} } } ]

                                                = [335.89 , 356.10]

Therefore, 96% confidence interval estimate for the mean monthly rent of all unmarried BYU students in winter 2018 is [335.89 , 356.10].

5 0
3 years ago
Can someone please help me with this?
Bad White [126]
I can’t see it sorry
7 0
3 years ago
Read 2 more answers
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