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fenix001 [56]
2 years ago
9

Given m∥n, find the value of x and y.

Mathematics
1 answer:
kupik [55]2 years ago
3 0
X=2
y=149
————-
Solve 8x+15=7x+17
alternate angles are equal
x=2

Substitute 2 in 8x+15 =31
180-31=149
y=149
Angles on a straight line add up to 180
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Y = 3x - 3| + 1<br> what is the domain and range?
Dimas [21]
Domain is the set of input values whilst range is the set of output values. i think 3x is your domain and -3 +1 is your domain
5 0
4 years ago
How much more expensive is it to buy a ticket in New york than in Atlanta?
gizmo_the_mogwai [7]

Answer:B


Step-by-step explanation:


6 0
3 years ago
Your car repair bill was $390. You were charged $215 for parts and $35 per hour for labor.
Zolol [24]

Answer:

5 hours

Step-by-step explanation:

subtract 390 by 215 to get 175 then divide 175/35 to get 5

5 0
3 years ago
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
Determine the slope of line whose equation is 5x+2y=10
Anni [7]
Move the x over to the other side
2y=10-5x
Divide both sides by 2 to get the y alone
y=-5/2x+5
Find m from the equation of a line (y=mx+b)
m=-5/2
Slope=-5/2
6 0
3 years ago
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