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34kurt
3 years ago
5

4⋅10 ​6 ​​ 4, dot, 10, start superscript, 6, end superscript is how many times as large as 1\cdot10^41⋅10 ​4 ​​ 1, dot, 10, star

t superscript, 4, end superscript?
Mathematics
1 answer:
lesya [120]3 years ago
6 0

The given problem is very confusing since it was copy pasted directly from the source so the equations look scrambled and plus it was one words. After my own translation, I believe the given numbers are:

4 ⋅ 10^6

and

1 ⋅ 10^4

The symbol ⋅ means that the two numbers are multiplied while the symbol ^ means an exponent of. Now we are asked to find how much the 1st number is larger than the 2nd number. To solve this, we simply have to divide the bigger number by the smaller number. Since 4 ⋅ 10^6 has bigger exponent than 1 ⋅ 10^4 then it is the bigger number.

Ratio = 4 ⋅ 10^6 / 1 ⋅ 10^4

Ratio = 4 ⋅ 10^2 = 400

Therefore 4 ⋅ 10^6 is 400 times bigger than 1 ⋅ 10^4.

 

Answer:

<span>400 times</span>

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What is the median in math
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In math, the median in a group of values is the value that is closest to the center of the group when the values are organized from least to greatest.

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Lina20 [59]

Answer: 120 beats per minute

Step-by-step explanation:

If you look at the plotted points on the graph, you can see that the rate is decreasing at a constant rate. As each week passes, the beats per minute rate drops by 10 beats. Therefore on Week 5, the heart rate will be 120 beats per minute.

4 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!

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4 years ago
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vredina [299]

Answer:

2.35%

Step-by-step explanation:

Mean number of months (M) = 39 months

Standard deviation (S) = 10 months

According to the 68-95-99.7 rule, 95% of the data is comprised within two standard deviations of the mean (39-20 to 39+20 months), while 99.7% of the data is comprised within two standard deviations of the mean (39-30 to 39+30 months).

Therefore, the percentage of cars still in service from 59 to 69 months is:

P_{59\ to\ 69}=\frac{P_{9\ to\ 69}-P_{19\ to\ 59}}{2} \\P_{59\ to\ 69}=\frac{99.7-95}{2}\\P_{59\ to\ 69}=2.35\%

The approximate percentage of cars that remain in service between 59 and 69 months is 2.35%.

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