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Ray Of Light [21]
3 years ago
15

4. Solve x -- 7 = 25. A 18 B 32

Mathematics
2 answers:
MAVERICK [17]3 years ago
5 0

b okay it b good

jwnsndm

skskxkmxxmcmc

jsjdkdjxkxxkxi

gregori [183]3 years ago
3 0

Answer:

B

Step-by-step explanation:

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23 degrees below 0, what is the answer please help
IrinaVladis [17]

Answer:

-23

Step-by-step explanation:

23 degrees below 0 is 0-23, or -23

Hope this helps plz mark brainliest :D

6 0
3 years ago
Read 2 more answers
An experiment was conducted to compare the use of iPads versus regular textbooks in teaching algebra to two classes of middle sc
lakkis [162]

Answer:

Step-by-step explanation:

Hello!

The objective is to determine if there is any difference between using iPads vs textbooks in teaching algebra.

Two middle school classes were selected, to eliminate any other source of variation, the same teacher taught both classes, and the materials were provided by the same author and publisher. After a month 10 students of each class were randomly selected and tested, their test scores were recorded:

X₁: test scores of students that used iPads to study.

n₁= 10

X[bar]₁= 86.8

S₁= 8.97

X₂: test scores of students that used regular textbooks to study.

n₂= 10

X[bar]₂= 79.5

S₂= 10.8

a.

H₀: μ₁=μ₂

H₁: μ₁≠μ₂

α:0.05

Assuming that both variables are normally distributed and the population variances are equal, the statistic to use is a Student t for two independent samples with pooled sample variance:

t_{H_0}= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{Sa*\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }

Sa^2= \frac{(n_1-1)S^2_1+(n_2-1)S^2_2}{n_1+n_2-2}

Sa^2= \frac{9*80.4609+9*116.64}{10+10-2} = 98.55

Sa= 9.93

t_{H_0}= \frac{86.8-79.5}{9.93\sqrt{\frac{1}{10} +\frac{1}{10} } } = 1.64

p-value: 0.118364

The p-value is greater than the significance level so the decision is to not reject the null hypothesis. This means that there is no significant evidence between the scores of the two groups.

b.

95% CI

(X[bar]-X[bar])±t_{n_1+n_2-2;1-\alpha /2}*Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} }

t_{n_1+n_2-2;1-\alpha /2}= t_{18;1.975}= 2.101

(86.8-79.5)±2.101*(9.93\sqrt{\frac{1}{10} +\frac{1}{10} })

[-2.03; 16.63]

With a 95% confidence level, you'd expect that the interval [-2.03; 16.63] would contain the difference between the mean scores of the two classes.

c.

Considering that the null hypothesis wasn't rejected and that at the same level the confidence interval includes the zero, we can affirm that the format of the teaching materials, digital or regular textbooks, has no significant effect on the scores of the students.

I hope it helps!

3 0
4 years ago
The function y=6+1.25x can be used to find the cost of joining an online music club and buying x songs. Which statement is true
krok68 [10]
B) Is the correct answer

We know it's B because it's a one off cost and isn't effected by x ( the amount of songs purchased )

4 0
4 years ago
£108 in the ratio 7:5
dybincka [34]

Answer:

45:63

Step-by-step explanation:

7+5=12

12u-->108

1u--> 9

5u-->45

7u-->63

8 0
3 years ago
Lines l and m are parallel lines cut by the transversal line t. Which angle is congruent to ∠7?
pav-90 [236]

Answer:

<3, <2 or <6

Step-by-step explanation:

6 0
3 years ago
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