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3 years ago

2) There are 40 boys and 60 girls in a class of students. What is the ratio of girls to students

Mathematics

2 answers:

ra1l [238]3 years ago

8 0

Answer:

60:100, 6/10, 3/5, 6 to 10, etc.

Step-by-step explanation:

You take the number of girls over total students which is boys + girls. Since there's 40 boys and 60 girls, it's 60 girls to 100 students which can be written in several ways.

Vikki [24]3 years ago

5 0

Answer:

60:100 / 3:5

Step-by-step explanation:

You first add the total number of students which is (40boys + 60girls) which gives us 100 students.

Then arrange the ratio of girls to students as per the question that is 60:100, reduce it to its lowest term that is dividing the ratio by 20, and finally got 3:5

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3 years ago

A college counselor is interested in estimating how many credits a student typically enrolls in each semester. The counselor dec

Ket [755]

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(a) The usual load is not 13 credits.

(b) The probability that a a student at this college takes 16 or more credits is 0.1093.

Step-by-step explanation:

According to the Central limit theorem, if a large sample (n ≥ 30) is selected from an unknown population then the sampling distribution of sample mean follows a Normal distribution.

The information provided is:

$Min.=8\\Q_{1}=13\\Median=14\\Mean=13.65\\SD=1.91\\Q_{3}=15\\Max.=18$

The sample size is, n = 100.

The sample size is large enough for estimating the population mean from the sample mean and the population standard deviation from the sample standard deviation.

So,

$\mu_{\bar x}=\bar x=13.65\\SE=\frac{s}{\sqrt{n}}=\frac{1.91}{\sqrt{100}}=0.191$

(a)

The null hypothesis is:

H₀: The usual load is 13 credits, i.e. μ = 13.

Assume that the significance level of the test is, α = 0.05.

Construct a (1 - α) % confidence interval for population mean to check the claim.

The (1 - α) % confidence interval for population mean is given by:

$CI=\bar x\pm z_{\alpha/2}\times SE$

For 5% level of significance the two tailed critical value of z is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Construct the 95% confidence interval as follows:

$CI=\bar x\pm z_{\alpha/2}\times SE\\=13.65\pm (1.96\times0.191)\\=13.65\pm0.3744\\=(13.2756, 14.0244)\\=(13.28, 14.02)$

As the null value, μ = 13 is not included in the 95% confidence interval the null hypothesis will be rejected.

Thus, it can be concluded that the usual load is not 13 credits.

(b)

Compute the probability that a a student at this college takes 16 or more credits as follows:

$P(X\geq 16)=P(\frac{X-\mu}{\sigma}\geq \frac{16-13.65}{1.91})\\=P(Z>1.23)\\=1-P(Z$

Thus, the probability that a a student at this college takes 16 or more credits is 0.1093.

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3 years ago

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