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2 years ago

Yesterday 5% of the 120 sixth graders at school were late. How many sixth graders were late?

Mathematics

1 answer:

attashe74 [19]2 years ago

6 0

Answer:

Step-by-step explanation:

5% of 120 were late

= $\frac{5}{100}$ × 120

= 0.05 × 120

= 6

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2 years ago

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3 years ago

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2 years ago

Find the least squares regression equation using the school year (in number of years after 2000) for the input variable and the

Ad libitum [116K]

Answer-

The least squares regression equation using the school year (in number of years after 2000) for the input variable and the average cost (in thousands of dollars) for the output variable is "y=0.937x+12.765" .

Solution-

The independent variable / input variable= x = Number of years after 2000 ( = year-2000)

The dependent variable / output variable = y = Average cost in thousands of dollars

(The table has been attached herewith.)

To find the regression equation for a group of (x,y) points,

We have to calculate the slope and y-intercept, then we can put those values in the equation y = mx + c ( Slope - Intercept formula)

We know that,

$Slope=m=\frac{N\sum xy -\sum x\sum y}{N\sum x^{2}- (\sum x)^{2}}$

Putting the values from the table,

$m=\frac{(8 \times 1020.206)-(52 \times 150.894)}{(8 \times 3810)-(52^2)}$

( ∵ Instead of 150,894 we have to put 150.894 as we have find the line for year and thousands of dollars )

$\Rightarrow m =\frac{8161.648-7846.488}{3040-2704} =\frac{315.16}{336}=0.937$

Now, for the y-intercept,

$c =\frac{\sum y -m(\sum x)}{N}$

Putting the values,

$c=\frac{150.894-(52 \times 0.937)}{8}=\frac{102.125}{8}=12.765$

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3 0

3 years ago

A sample of size 200 will be taken at random from an infinite population. given that the population proportion is 0.60, the prob

Musya8 [376]

Let p be the population proportion. 
We have p=0.60, n=200 and we are asked to find P(^p<0.58). 
The thumb of the rule is since n*p = 200*0.60 and n*(1-p)= 200*(1-0.60) = 80 are both at least greater than 5, then n is considered to be large and hence the sampling distribution of sample proportion-^p will follow the z standard normal distribution. Hence this sampling distribution will have the mean of all sample proportions- U^p = p = 0.60 and the standard deviation of all sample proportions- δ^p = √[p*(1-p)/n] = √[0.60*(1-0.60)/200] = √0.0012.
So, the probability that the sample proportion is less than 0.58
= P(^p<0.58)
= P{[(^p-U^p)/√[p*(1-p)/n]<[(0.58-0.60)/√0...
= P(z<-0.58)
= P(z<0) - P(-0.58<z<0)
= 0.5 - 0.2190
= 0.281
So, there is 0.281 or 28.1% probability that the sample proportion is less than 0.58.

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3 years ago