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3 years ago

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The ratio of blue tacks to red tacks on the bulletin board is 3 to 5. There are 50 red tacks. How many blue tacks are on the boa

rd?

1 answer:

Aleksandr [31]3 years ago

8 0

3/5 times 50/1= 30 blue tacks

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Greg budgeted carefully and by the time he turned 30 his retirement account balance was $22,000. If that amount grew at a rate o

Ratling [72]

Answer:

$618,253.61

Step-by-step explanation:

30 to 65 is 35 years

22000 × (1 + 10/100)³⁵

22000 × 1.1³⁵

618253.6107

6 0

3 years ago

Can someone please help with this question

maksim [4K]

Answer:

X = 7.2

Step-by-step explanation:

Hello

For the triangle of right side, using Pytagoras:

(2.5)^2 + b^2 = 6^2

6.25 + b^2 = 36

b^2 = 36 - 6.25

b^2 = 29.75

b = 5.45

Now we can to find X using again Pytagoras:

3^2 + (5.45)^2 = x^2

9 + 29.75 = x^2

38.75 = x^2

X = 7.2

Best regards

5 0

3 years ago

Which one doesn’t belong

aniked [119]

Answer:

x+5=5

Step-by-step explanation:

This is an equation where you are actually solving it. The others are expressions.

3 0

3 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

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

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

<u>95% confidence interval for</u> ( $\mu_1-\mu_2$ ) =

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

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

= [-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

Please answer one question asap <br>A(-4,3) B(2,-1) D(-4,-1) C(2,-5)

Leni [432]

Answer:

Step-by-step explanation:

Vertices of the given quadrilateral are A(-4, 3), B(2, -1), C(2, -5) and D(-4, -1)

Since, slope of a line passing through two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by,

Slope = $\frac{y_2-y_1}{x_2-x_1}$

Slope of AB = $\frac{3+1}{-4-2}$

= $-\frac{2}{3}$

Slope of AD = $\frac{3+1}{-4+4}$

= Not defined (Parallel to y-axis)

Slope of DC = $\frac{-5+1}{2+4}$

= $-\frac{2}{3}$

Slope of BC = $\frac{-1+5}{2-2}$

= Not defined (Parallel to y-axis)

Slope of AB = slope of DC = $-\frac{2}{3}$

Slope of BC = slope of AD = Not defined (parallel to y-axis)

As per property of a parallelogram,

"Opposite sides of a parallelogram are parallel and equal in measure"

Therefore, ABCD is a parallelogram.

8 0

3 years ago