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2 years ago

8

Mr. Malone made a graph to represent the time his students spent studying for their tests throughout the unit and their actual t

est scores Use the two points that are marked on the line of best fit to answer the following questions .

1 answer:

denis-greek [22]2 years ago

8 0

Answer:

a. 4/3

b. Test scores in percentage

c. y=4/3x +50

d. 78

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Each homeroom must send two student representatives to each student council meeting. Mrs. Lincoln has 15 girls and 10 boys in he

Ierofanga [76]

Answer:

1/4

Step-by-step explanation:

Mrs. Lincoln has 15 girls and 10 boys in her homeroom. She randomly selects two students. The total number of students is 25.

The probability of choosing one girl is:

15 / 25 = 3 / 5

Now, 24 students are left.

The probability of choosing one boy is:

10 / 24 = 5 / 12

Therefore, the probability of choosing one girl and one boy is:

3 / 5 * 5 / 12

= 3 / 12

= 1/4

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2 years ago

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Step-by-step explanation:

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2 years ago

Find the value of the expression 2ab/c when a = -1, b = -4, and c= 2

motikmotik

Answer:

4

Step-by-step explanation:

Given

$\frac{2ab}{c}$ , substitute the given values into the expression

= $\frac{2(-1)(-4)}{2}$

= $\frac{2(4)}{2}$

= $\frac{8}{2}$

= 4

4 0

2 years ago

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Construct a 99% confidence interval for the population mean, mu. Assume the population has a normal distribution. A group of 1

Zarrin [17]

Answer:

99% confidence interval for the population mean is [19.891 , 24.909].

Step-by-step explanation:

We are given that a group of 19 randomly selected students has a mean age of 22.4 years with a standard deviation of 3.8 years.

Assuming the population has a normal distribution.

Firstly, the pivotal quantity for 99% confidence interval for the population mean is given by;

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

where, $\bar X$ = sample mean age of selected students = 22.4 years

s = sample standard deviation = 3.8 years

n = sample of students = 19

$\mu$ = population mean

<em>Here for constructing 99% confidence interval we have used t statistics because we don't know about population standard deviation.</em>

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-2.878 < $t_1_8$ < 2.878) = 0.99 {As the critical value of t at 18 degree of

freedom are -2.878 & 2.878 with P = 0.5%}

P(-2.878 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 2.878) = 0.99

P( $-2.878 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.878 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

P( $\bar X -2.878 \times {\frac{s}{\sqrt{n} }$ < $\mu$ < $\bar X +2.878 \times {\frac{s}{\sqrt{n} }$ ) = 0.99

<u>99% confidence interval for</u> $\mu$ = [ $\bar X -2.878 \times {\frac{s}{\sqrt{n} }$ , $\bar X +2.878 \times {\frac{s}{\sqrt{n} }$ ]

= [ $22.4 -2.878 \times {\frac{3.8}{\sqrt{19} }$ , $22.4 +2.878 \times {\frac{3.8}{\sqrt{19} }$ ]

= [19.891 , 24.909]

Therefore, 99% confidence interval for the population mean is [19.891 , 24.909].

6 0

2 years ago

Help once again !!!!

alekssr [168]

The answer is a rotation

6 0

2 years ago

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