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damaskus [11]
3 years ago
8

If A:(1,4,7,8) and b:(4,6,8,9)and c:​

Mathematics
1 answer:
Usimov [2.4K]3 years ago
5 0
Incomplete question please add a image
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One-third of the class grew peas, one-third grew carrots, and one-third grew beans. The class consisted of 63 students of which
olga_2 [115]

The number of boys and girls assigned to each team is 6 and 15 respectively.

<h3>Fraction</h3>
  • Number of students in class = 63

  • Number of students who grows peas = 1/3 × 63

= 63/3

= 21 students

  • Number of students who grows carrots = 21
  • Number of students who grows beans = 21

Number of girls = 5/7 × 63

= 315/7

Number of girls = 45 students

Number of boys = 63 - 45 = 18 students

If the three groups have equal composition of boys and girls, then,

Number of boys and girls in each group is 6 and 15 respectively.

Learn more about fraction:

brainly.com/question/11562149

#SPJ1

7 0
2 years ago
Natasha has a gross income of $66,429. She can make adjustments of $14,490 for business losses, $3,584 for business expenses, an
Alecsey [184]
Your answer is D. I just answered this question
6 0
3 years ago
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Make -4 a fraction ​
stiv31 [10]

Answer:

4/1

Step-by-step explanation:

like 1 is like 1 whole

4 0
3 years ago
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the area of a rectangular swimming pool is 10x^2-19-15 the length is 5x+3 .what is the width of the pool?
irga5000 [103]
2x-5 is the answer
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3 years ago
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Brainliest + 20 pts to whoever helps pls!!
Blizzard [7]

Answer:

Correct option is (a). 60.391 and 101.879; because the test statistic is in a critical region, the test rejects the null hypothesis.

Step-by-step explanation:

A Chi-square test for population variance is used to perform this test.

The standard deviation is, 2.0 minutes.

Then the variance is, 4.0 minutes.

The hypothesis for this test is:

<em>H₀</em>: The population variance of all commute times is equal to 4.0 minutes, i.e. <em>σ² </em>= 4.

<em>Hₐ</em>: The population variance of all commute times is not equal to 4.0 minutes, i.e. <em>σ² </em>≠ 4.

The test statistic is:

\chi^{2}_{calc.}=\frac{(n-1)s^{2}}{\sigma^{2}}

The critical region of this test is defined as:

Reject <em>H₀</em> if \chi^{2}_{calc.} or \chi^{2}_{calc.}>\chi^{2}_{(1-\alpha /2), (n-1)}.

The degrees of freedom is:

n-1=81-1=80

Compute the critical from a Chi-square table.

\chi^{2}_{\alpha /2, (n-1)}=\chi^{2}_{0.05, 80}=101.879\\\chi^{2}_{(1-\alpha /2), (n-1)}=\chi^{2}_{0.95, 80}=60.391\\

The test statistic value is, \chi^{2}_{calc.}=105.8.

\chi^{2}_{calc.}=105.8 > \chi^{2}_{0.05, 80}=101.879

The null hypothesis is rejected because the test statistic is in the critical region.

Thus, the correct option is (a).

6 0
3 years ago
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