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jeka94
4 years ago
15

It takes 6 painters 20 hours working together to paint 15 rooms. How long will it take 1 painter to paint 1 room working alone a

t the same rate?
Mathematics
1 answer:
NISA [10]4 years ago
8 0
It will take about 22 minutes because 20 hours divided by 15 rooms equals 1 hour and 33 minutes for one room divided by the six painters equals 22 minutes
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Find a coefficient of x in -7x, and a coefficient of 9/4y.
garik1379 [7]

Answer:9

Step-by-step explanation:

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3 years ago
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Consider each of the following relationships. Which school shows the greatest rate of change between male and female students?
ratelena [41]

Answer:

The school C shows the greatest rate of change between male and female students

Step-by-step explanation:

Let

x -----> the number of female students

y ----> the number of male students

we know that

The ratio of male students to female students is

\frac{y}{x}

we have

<em>School A</em>

\frac{y}{x}=\frac{4}{1}=4

<em>School B</em>

\frac{y}{x}=\frac{4}{3}=1.33 ----> the slope of the linear equation

<em>School C</em>

\frac{y}{x}=\frac{1,260}{280}=4.5

so

4.5 > 4 > 1.33

therefore

The school C shows the greatest rate of change between male and female students

7 0
3 years ago
16)
elixir [45]

$78

The difference between choice A and B is 5 meals and $80 so 1 meal is $16. So we can subtract $16 from $250 and we get $234 for 3 nights. 1 night is $78

8 0
3 years ago
Suppose an investment of $2,300 doubles in value every decade. The function F(X)= 2300 * 2^X gives the value of the investment a
kotykmax [81]

You hit the right target!

\[\Huge\color{}\checkmark\]

It is 9200

Just to be sure, substitute with x = 2 in the equation.

8 0
3 years ago
The International Air Transport Association surveys business travelers to develop quality ratings for transatlantic gateway airp
motikmotik

Answer:

The 95% confidence interval for the population mean rating is (5.73, 6.95).

Step-by-step explanation:

We start by calculating the mean and standard deviation of the sample:

M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{50}(6+4+6+. . .+6)\\\\\\M=\dfrac{317}{50}\\\\\\M=6.34\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{49}((6-6.34)^2+(4-6.34)^2+(6-6.34)^2+. . . +(6-6.34)^2)}\\\\\\s=\sqrt{\dfrac{229.22}{49}}\\\\\\s=\sqrt{4.68}=2.16\\\\\\

We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=6.34.

The sample size is N=50.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

s_M=\dfrac{s}{\sqrt{N}}=\dfrac{2.16}{\sqrt{50}}=\dfrac{2.16}{7.071}=0.305

The degrees of freedom for this sample size are:

df=n-1=50-1=49

The t-value for a 95% confidence interval and 49 degrees of freedom is t=2.01.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_M=2.01 \cdot 0.305=0.61

Then, the lower and upper bounds of the confidence interval are:

LL=M-t \cdot s_M = 6.34-0.61=5.73\\\\UL=M+t \cdot s_M = 6.34+0.61=6.95

The 95% confidence interval for the mean is (5.73, 6.95).

4 0
3 years ago
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