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Serjik [45]
3 years ago
12

Find the product. 5x ^2 · 2y ^2

Mathematics
1 answer:
siniylev [52]3 years ago
3 0
The answer should be 10x²y²
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Explain why thirteen
insens350 [35]

Answer:

thirteen hundredths=13/100

one hundred thirty thousandths=130/1000

cancelling by 10 this is =13/100

hence the question

Step-by-step explanation:

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4 years ago
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A shop keeper bought 26 apples from a fruit vendor for $37.70. How much did each apple cost?
Semenov [28]
You have to divide 37.70. By 26
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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
Coach Kasongo wants to know how many of his swimmers have times in each of four time ranges. He times each swimmer and records t
alina1380 [7]

Answer: D

Step-by-step explanation:

6 0
3 years ago
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The following is an arithmetic series.<br> 31+33.5+36+38.5+...<br> O True<br> O False
Aleks04 [339]

9514 1404 393

Answer:

  True

Step-by-step explanation:

It is the sum of numbers having a common difference. The fact that it is a sum makes it a series (as opposed to a sequence, which is just a list of numbers). The common difference (of 2.5) makes it an arithmetic series.

True, the sum is an arithmetic series.

3 0
3 years ago
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