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3 years ago

There are 27 students in a class. If 1/3 of the students are girls, how many boys are in the class?

Mathematics

2 answers:

Butoxors [25]3 years ago

7 0

Answer:

27x 1/3= 9

9 boys are there in the class

Alex73 [517]3 years ago

7 0

18 boys are in the class. If 1/3 of the class is girls (9 girls) then we can do 27-9=18.

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Find the area of the circle.

juin [17]

Answer:

See Explanation Below

Step-by-step explanation:

Your question isn't clear; However, I'll solve in two ways

1. Diameter = 14 cm

2. Radius = 14 cm

When Diameter = 14 cm

Given

$Area = \pi r^2$

$Diameter = 14cm$

Required

Calculate the area of the circle

First, the radius has to be solved;

$Radius = \frac{1}{2} Diameter$

$Radius = \frac{1}{2} * 14cm$

$Radius = 7cm$

Using the given formula;

The area is as follows

$Area = \pi r^2$

$Area = 3.14 * 7^2$

$Area = 3.14 * 49$

$Area = 153.86cm^2$

When Radius = 14 cm

Given

$Area = \pi r^2$

$Radius = 14cm$

Required

Calculate the area of the circle

Using the given formula;

The area is as follows

$Area = \pi r^2$

$Area = 3.14 * 14^2$

$Area = 3.14 * 196$

$Area = 615.44cm^2$

7 0

3 years ago

Can you the nearest hundredth to 1.2983?

ozzi

I think so it’s 1.30

3 0

2 years ago

Read 2 more answers

I need help with the last two !!!

Dmitriy789 [7]

Step-by-step explanation:

G(x) = - x^2 + 10x

a + G(a) = a - a^2 + 10a = - a^2 + 11a or 11a - a^2 or a(11 - a)

1/G(a) = 1/(10a - a^2) = 1/a(10 - a)

6 0

2 years ago

Fast-food restaurants spend quite a bit of time studying the amount of time cars spend in their drive-throughs. Certainly, the f

melisa1 [442]

Answer:

For the 99th percentile, we have X = 206 seconds.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 138.5, \sigma = 29$

99th percentile:

Value of X when Z has a pvalue of 0.99. So we use $Z = 2.325$

$Z = \frac{X - \mu}{\sigma}$

$2.325 = \frac{X - 138.5}{29}$

$X - 138.5 = 29*2.325$

$X = 205.92 = 206$

For the 99th percentile, we have X = 206 seconds.

4 0

3 years ago

Three populations have proportions 0.1, 0.3, and 0.5. We select random samples of the size n from these populations. Only two of

IRINA_888 [86]

Answer:

(1) A Normal approximation to binomial can be applied for population 1, if n = 100.

(2) A Normal approximation to binomial can be applied for population 2, if n = 100, 50 and 40.

(3) A Normal approximation to binomial can be applied for population 2, if n = 100, 50, 40 and 20.

Step-by-step explanation:

Consider a random variable X following a Binomial distribution with parameters n and p.

If the sample selected is too large and the probability of success is close to 0.50 a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

np ≥ 10
n(1 - p) ≥ 10

The three populations has the following proportions:

p₁ = 0.10

p₂ = 0.30

p₃ = 0.50

(1)

Check the Normal approximation conditions for population 1, for all the provided n as follows:

$n_{a}p_{1}=10\times 0.10=1$

Thus, a Normal approximation to binomial can be applied for population 1, if n = 100.

(2)

Check the Normal approximation conditions for population 2, for all the provided n as follows:

$n_{a}p_{1}=10\times 0.30=310\\\\n_{c}p_{1}=50\times 0.30=15>10\\\\n_{d}p_{1}=40\times 0.10=12>10\\\\n_{e}p_{1}=20\times 0.10=6$

Thus, a Normal approximation to binomial can be applied for population 2, if n = 100, 50 and 40.

(3)

Check the Normal approximation conditions for population 3, for all the provided n as follows:

$n_{a}p_{1}=10\times 0.50=510\\\\n_{c}p_{1}=50\times 0.50=25>10\\\\n_{d}p_{1}=40\times 0.50=20>10\\\\n_{e}p_{1}=20\times 0.10=10=10$

Thus, a Normal approximation to binomial can be applied for population 2, if n = 100, 50, 40 and 20.

8 0

3 years ago