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3 years ago

You have to download question

Mathematics

1 answer:

solmaris [256]3 years ago

4 0

Answer:

B) Mode doesn't change.

Step-by-step explanation:

Given:

A set of data with 16 being added to the set.

The mode of a given set of data is the data that is repeated the most number of times.

Here, 19 is repeated 3 times, so the mode is 19.

Now, if a number other than that of 19 being added, doesn't affect the mode of the data set as 19 will still be 3 times and adding 16 will not increase its number in the set.

So, the correct option is B.

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Help please Hurry!!!!

Alex787 [66]

C I know I took test

7 0

2 years ago

A university wants to compare out-of-state applicants' mean SAT math scores (?1) to in-state applicants' mean SAT math scores (?

nordsb [41]

Answer:

d. Yes, because the confidence interval does not contain zero.

Step-by-step explanation:

We are given that the university looks at 35 in-state applicants and 35 out-of-state applicants. The mean SAT math score for in-state applicants was 540, with a standard deviation of 20.

The mean SAT math score for out-of-state applicants was 555, with a standard deviation of 25.

Firstly, the Pivotal quantity for 95% confidence interval for the difference between the population means is given by;

P.Q. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ~ $t__n__1-_n__2-2$

where, $\bar X_1$ = sample mean SAT math score for in-state applicants = 540

$\bar X_2$ = sample mean SAT math score for out-of-state applicants = 555

$s_1$ = sample standard deviation for in-state applicants = 20

$s_2$ = sample standard deviation for out-of-state applicants = 25

$n_1$ = sample of in-state applicants = 35

$n_2$ = sample of out-of-state applicants = 35

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(35-1)\times 20^{2} +(35-1)\times 25^{2} }{35+35-2} }$ = 22.64

Here for constructing 95% confidence interval we have used Two-sample t test statistics.

So, 95% confidence interval for the difference between population means ( $\mu_1-\mu_2$ ) is ;

P(-1.997 < $t_6_8$ < 1.997) = 0.95 {As the critical value of t at 68 degree

of freedom are -1.997 & 1.997 with P = 2.5%}

P(-1.997 < $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < 1.997) = 0.95

P( $-1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < ${(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}$ < $1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ) = 0.95

P( $(\bar X_1-\bar X_2)-1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < ( $\mu_1-\mu_2$ ) < $(\bar X_1-\bar X_2)+1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ) = 0.95

95% confidence interval for ( $\mu_1-\mu_2$ ) =

[ $(\bar X_1-\bar X_2)-1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ , $(\bar X_1-\bar X_2)+1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ]

=[ $(540-555)-1.997 \times {22.64 \times \sqrt{\frac{1}{35} +\frac{1}{35} } }$ , $(540-555)+1.997 \times {22.64 \times \sqrt{\frac{1}{35} +\frac{1}{35} } }$ ]

= [-25.81 , -4.19]

Therefore, 95% confidence interval for the difference between population means SAT math score for in-state and out-of-state applicants is [-25.81 , -4.19].

This means that the mean SAT math scores for in-state students and out-of-state students differ because the confidence interval does not contain zero.

So, option d is correct as Yes, because the confidence interval does not contain zero.

6 0

3 years ago

Complete the statement to describe the expression (a+b+c)(d+e+f)

kherson [118]

Answer:

The given expression consists of 2 factors, and each factor contains 3 terms.

Step-by-step explanation:

Given the expression (a+b+c)(d+e+f)

Factors are parts of expression that are connected by multiplication. we are multiplying (a+b+c) and (d+e+f) so we said (a+b+c) and (d+e+f) are factors of the expression.

Hence, there are 2 factors in given expression.

A mathematical expression contains numbers, variables and operators joined by addition, subtraction, multiplication, and division. The parts of the expression that are connected with addition and subtraction are known as terms.

In each factor (a+b+c) and (d+e+f) three terms are connected by addition. Hence, there are 3 terms in both the factors.

7 0

3 years ago

Write the first 4 terms for f (1) = 3, f (n) = [f(n-1)]

Alexxx [7]

$f(1)=3\\\\f(n)=[f(n-1)]\\\\f(2)=[f(2-1)]=[f(1)]=[3]=3\\\\f(3)=[f(3-1)]=[f(2)]=[3]=3\\\\f(4)=[f(4-1)]=[f(3)]=[3]=3\\\\f(5)=[f(5-1)]=[f(4)]=[3]=3$

4 0

3 years ago

20% of a is 11. what is a?

Oksana_A [137]

Answer:

a=55

Step-by-step explanation:

20%/100=0.2

a=11/0.2

a=55

5 0

3 years ago