PLEASE HELP ASAP WILL GIVE BRAINLEIST!!!

In a school, two-fifths of the pupils are girls. how many pupils are there if 96 are boys

NikAS [45]

38.4 is the answer i think because u divide 96 by 5 then times it by 2 to get two fifths

5 0

3 years ago

Questions (no partial grades if you don't show your work) 1. In a group of 6 boys and 4 girls, four children are to be selected.

mariarad [96]

Answer:

Total number of ways will be 209

Step-by-step explanation:

There are 6 boys and 4 girls in a group and 4 children are to be selected.

We have to find the number of ways that 4 children can be selected if at least one boy must be in the group of 4.

So the groups can be arranged as

(1 Boy + 3 girls), (2 Boy + 2 girls), (3 Boys + 1 girl), (4 boys)

Now we will find the combinations in which these arrangements can be done.

1 Boy and 3 girls = $^{6}C_{1}\times^{4}C_{3}=6\times4$ =24

2 Boy and 2 girls= $^{6}C_{2}\times^{4}C_{2}=\frac{6!}{4!\times2!}\times\frac{4!}{2!\times2!}=15\times6=90$

3 Boys and 1 girl = $^{6}C_{3}\times^{4}C_{1}=\frac{6!}{4!\times2!}\times\frac{4!}{3!}=\frac{6\times5\times4}{3 \times2} \times4=80$

4 Boys = $^{6}C_{4}=\frac{6!}{4!\times2!} =\frac{6\times 5}{2\times1}=15$

Now total number of ways = 24 + 90 + 80 + 15 = 209

7 0

3 years ago

Find mike average monthly income based on these weekly incomes

7nadin3 [17]

Answer:

Mike's average monthly salary is $ 1,814.

Step-by-step explanation:

To determine Mike's average monthly salary, taking into account that he earns $ 284, $ 292, $ 418, $ 350, and $ 470 weekly, these amounts must be added for the 5-week duration of the month through the following calculation:

284 + 292 + 418 + 350 + 470 = X

1814 = X

Thus, given that this sum totals $ 1,814, Mike's average monthly salary is $ 1,814.

5 0

3 years ago

What are the zeros of this function?

BabaBlast [244]

Answer:

x = 0 and x = 6

Step-by-step explanation:

See the graph attached to this question.

It is clear from the graph that the curve touches x-axis i.e. y becomes zero at points x = 0 and x = 6

Therefore, the give parabola has zeros at x = 0 and x = 6. (Answer)

Zeros of a function y = f(x) are determined from the equation f(x) = 0 and solving it we will get the x-values which are the zeros of the function.

3 0

3 years ago

A study by the National Athletic Trainers Association surveyed random samples of 1679 high school freshmen and 1366 high school

nekit [7.7K]

Answer:

We conclude that there is no significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids.

Step-by-step explanation:

We are given that a study by the National Athletic Trainers Association surveyed random samples of 1679 high school freshmen and 1366 high school seniors in Illinois.

Results showed that 34 of the freshmen and 24 of the seniors had used anabolic steroids.

Let $p_1$ = proportion of Illinois high school freshmen who have used anabolic steroids.

$p_2$ = proportion of Illinois high school seniors who have used anabolic steroids.

SO, Null Hypothesis, $H_0$ : $p_1=p_2$ {means that there is no significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids}

Alternate Hypothesis, $H_A$ : $p_1\neq p_2$ {means that there is a significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids}

The test statistics that would be used here Two-sample z test for proportions;

T.S. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of high school freshmen who have used anabolic steroids = $\frac{34}{1679}$ = 0.0203

$\hat p_2$ = sample proportion of high school seniors who have used anabolic steroids = $\frac{24}{1366}$ = 0.0176

$n_1$ = sample of high school freshmen = 1679

$n_2$ = sample of high school seniors = 1366

So, the test statistics = $\frac{(0.0203-0.0176)-(0)}{\sqrt{\frac{0.0203(1-0.0203)}{1679}+\frac{0.0176(1-0.0176)}{1366} } }$

= 0.545

The value of z test statistics is 0.545.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistic lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that there is no significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids.

3 0

3 years ago