Mr. Salazar assigns the same number of homework questions each school day. The ratio table below shows the total

In a normally distributed data set a mean of 55 where 95% of the data fall between 47.4 and 62.6, what would be the standard dev

tresset_1 [31]

Answer:

The standard deviation of that data set is 3.8

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 55

95% of the data fall between 47.4 and 62.6. This means that 47.4 is 2 standard deviations below the mean and 62.6 is two standard deviations above the mean.

Using one of these points.

55 + 2sd = 62.6

2sd = 7.6

sd = 7.6/2

sd = 3.8

The standard deviation of that data set is 3.8

6 0

2 years ago

Which statements are true for the functions g(x) = x2 and h(x) = –x2 ? Check all that apply.

Andre45 [30]

If x=0 then they have same value

1st and 2nd options are out

for x=-1
g(-1)=1
h(-1)=-1
3rd is true

4th
false

for all values except zero, g(x)>h(x)

correct ones are

g(x) > h(x) for x = -1.
For positive values of x, g(x) > h(x).
For negative values of x, g(x) > h(x).

6 0

2 years ago

Read 2 more answers

8.5b−3.4∙(13a−3.2b)+a

Nesterboy [21]

Answer:

8.5b - 3.4(13a - 3.2b) + a = 19.4b - 43.2a

Step-by-step explanation:

It is a simple mathematical problem with multiple like terms. We can solve it by applying basic mathematical rules of multiplication and addition/subtration.

8.5b - 3.4(13a - 3.2b) + a

= 8.5b - 3.4*13a -3.4*(-3.2b) + a

= 8.5b - 3.4*13a + 3.4*3.2b + a

= 8.5b - 44.2a + 10.88 b + a

Now, only like terms can be added to each other

= (8.5b + 10.9b) + (a - 44.2a)

= 19.4b + (-43.2a)

= 19.4b - 43.2a

8 0

2 years ago

Identify the 12th term of the arithmetic sequence in which a7 = 40 and a18 = 106.

Lady_Fox [76]

we are given

airthematic sequence

we can use nth term formula

$a_n=a_1+(n-1)d$

n is number of terms

an is nth term

d is common difference

a1 is first term

We are given

a7=40

we can use it

$a_7=a_1+(7-1)d$

$40=a_1+6d$

$a_1+6d=40$

a18 = 106

$a_1_8=a_1+(18-1)d$

$a_1+17d=106$

we can subtract both equations

$a_1+17d-a_1-6d=106-40$

$11d=66$

$d=6$

now, we can find a1

$a_1+6d=40$

we can plug d=6

$a_1+6*6=40$

$a_1=4$

12th term:

$a_1_2=a_1+(12-1)d$

now, we can plug values

$a_1_2=4+(12-1)*6$

$a_1_2=4+66$

$a_1_2=70$

so, 12th term is 70...........Answer

3 0

2 years ago

A researcher gathers data on the length of essays (number of lines) and the SAT scores received for a sample of students enrol

LenaWriter [7]

Answer:

(C) The slope of the regression equation is not significantly different from zero

Step-by-step explanation:

Let's suppose that we have the following linear model:

$y= \beta_o +\beta_1 X$

Where Y is the dependent variable and X the independent variable. $\beta_0$ represent the intercept and $\beta_1$ the slope.

In order to estimate the coefficients $\beta_0 ,\beta_1$ we can use least squares estimation.

If we are interested in analyze if we have a significant relationship between the dependent and the independent variable we can use the following system of hypothesis:

Null Hypothesis: $\beta_1 = 0$

Alternative hypothesis: $\beta_1 \neq 0$

Or in other wouds we want to check is our slope is significant.

In order to conduct this test we are assuming the following conditions:

a) We have linear relationship between Y and X

b) We have the same probability distribution for the variable Y with the same deviation for each value of the independent variable

c) We assume that the Y values are independent and the distribution of Y is normal

The significance level is provided and on this case is $\alpha=0.05$

The standard error for the slope is given by this formula:

$SE_{\beta_1}=\frac{\sqrt{\frac{\sum (y_i -\hat y_i)^2}{n-2}}}{\sqrt{\sum (X_i -\bar X)^2}}$

Th degrees of freedom for a linear regression is given by $df=n-2$ since we need to estimate the value for the slope and the intercept.

In order to test the hypothesis the statistic is given by:

$t=\frac{\hat \beta_1}{SE_{\beta_1}}$

The confidence interval for the slope would be given by this formula:

$\hat \beta_1 + t_{n-2, \alpha/2} \frac{\sqrt{\frac{\sum (y_i -\hat y_i)^2}{n-2}}}{\sqrt{\sum (X_i -\bar X)^2}}$

And using the last formula we got that the confidence interval for the slope coefficient is given by:

$-0.88 < \beta_1$

IF we analyze the confidence interval contains the value 0. So we can conclude that we don't have a significant effect of the slope on this case at 5% of significance. And the best option would be:

(C) The slope of the regression equation is not significantly different from zero

6 0

2 years ago