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Licemer1 [7]
3 years ago
8

8-x+4-2-8x= What does x equal

Mathematics
1 answer:
goldfiish [28.3K]3 years ago
3 0

Answer:

-9x + 10

Step-by-step explanation:

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9n - 8d<br> (n = 3 and d = 2
leva [86]

Answer:

11

Step-by-step explanation:

9(3)-8(2)

27-16

11

3 0
2 years ago
a class consists of 10 boys and 5 girls. how many ways can a committee of 3 boys and 2 girls be selected
hammer [34]

Answer:

<em>i dont know the answer dude or ladies</em>

Step-by-step explanation:

\

8 0
3 years ago
Solve for x
Ulleksa [173]

Answer:

C. x = 1

Step-by-step explanation:

To solve this problem the best way is to use the quadratic formula. First make everything equal to 0 by adding 3 to both sides.

3x^2 - 6x + 3 = 0

The quadratic formula is: \frac{-b\pm\sqrt{b^2-4ac} }{2a}. Substitute the values of a, b, and c into the formula. Keep in mind:

  • a = 3
  • b = -6
  • c = 3

\frac{-(-6)\pm\sqrt{(-6)^2-4(3)(3)} }{2(3)} \rightarrow \frac{6\pm\sqrt{36-4(9)} }{6} \rightarrow \frac{6\pm\sqrt{0} }{6} \rightarrow \frac{1\pm0}{1} \rightarrow 1

Therefore, x = 1.

8 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
The test scores for a math class are shown below.
Nady [450]

Answer:

c. 4.7

Step-by-step explanation:

first, you find the mean of data set:

83 + 85 + 82 + 93 + 83 + 84 + 95 + 87 + 86 + 94 = 872/10 = 87.2

then, you take each number and subtract the mean and square it:

(87.2 - 83)² = (4.2)² = 17.64

(87.2 - 85)² = (2.2)² = 4.84

(87.2 - 82)² = (5.2)² = 27.04

(93 - 87.2)² = (5.8)² = 33.64

(87.2 - 83)² = (4.2)² = 17.64

(87.2 - 84)² = (3.2)² = 10.24

(95 - 87.2)² = (7.8)² = 60.84

(87.2 - 87)² = (0.2)² = 0.04

(87.2 - 86)² = (1.2)² = 1.44

(94 - 87.2)² = (6.8)² = 46.24

now, find the mean of the squared numbers:

17.64 + 4.84 + 27.04 + 33.64 + 17.64 + 10.24 + 60.84 + 0.04 + 1.44 + 46.24 = 219.6/10 = 21.96

then, take the square root of the number you just got:

√21.96 = 4.68

and last but not least, round to the nearest tenth: 4.7

6 0
3 years ago
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