I bought a new bike! It was originally $600 but I had a 25% coupon. What

FiftyFifty adult men in the United States are randomly selected and measured for their body mass index (BMI). Based on that sa

Fynjy0 [20]

Answer: $(23.3,\ 29.7)$

Step-by-step explanation:

The confidence interval for population mean is given by :-

$\overline{x}\pm ME.$ , where $\overline{x}$ is the sample mean and ME is the margin of error .

Given : The sample mean : $\overline{x}=26.5$

Margin of error : $ME=3.2$

Then , the range of values (confidence interval) likely to contain the true value of the population parameter will be :-

$26.5\pm 3.2=(26.5-3.2,\ 26.5+3.2)=(23.3,\ 29.7)$

Hence, the range of values likely to contain the true value of the population parameter = $(23.3,\ 29.7)$

3 0

3 years ago

Read 2 more answers

An animal shelter spends $3.00 per day to care for each bird and $8.50 per day to care for each cat. Kayla noticed that the shel

Lady bird [3.3K]

$123 = ($3b + $8.50c)

b + c = 30

b = 24 birds ($72)
c = 6 cats ($51)

51+72=123

There were 24 birds at the shelter on Tuesday.

8 0

3 years ago

Read 2 more answers

The inverse of an exponential function is the logarithmic function. True or False?

patriot [66]

For one-to-one functions, the inverse of an exponential function is a logarithmic function and the inverse of a logarithmic function is an exponential function.
For example:
$f(x)=e^{x}$
$f^{-1} (x)=ln\ x$

$f(x)=ln\ x$
$f^{-1} x=e^{x}$

5 0

3 years ago

A person purchased a slot machine and tested it by playing it 1,137 times. There are 10 different categories of outcomes, includ

ivann1987 [24]

Answer:

$p_v= P(\chi^2_{9}>11.517)=0.2419$

And on this case if we see the significance level given $\alpha=0.1$ we see that $p_v>alpha$ so we fail to reject the null hypothesis that the observed outcomes agree with the expected frequencies at 10% of significance.

Step-by-step explanation:

A chi-square goodness of fit test determines if a sample data obtained fit to a specified population.

$p_v$ represent the p value for the test

O= obserbed values

E= expected values

The system of hypothesis for this case are:

Null hypothesis: $O_i = E_i[/tex[Alternative hypothesis: [tex]O_i \neq E_i$

The statistic to check the hypothesis is given by:

$\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

On this case after calculate the statistic they got: $\chi^2 = 11.517$

And in order to calculate the p value we need to find first the degrees of freedom given by:

$df=n-1=10-1=9$ , where k represent the number of levels (on this cas we have 10 categories)

And in order to calculate the p value we need to calculate the following probability:

$p_v= P(\chi^2_{9}>11.517)=0.2419$

And on this case if we see the significance level given $\alpha=0.1$ we see that $p_v>alpha$ so we fail to reject the null hypothesis that the observed outcomes agree with the expected frequencies at 10% of significance.

6 0

4 years ago

Simplify fraction 4/28 and 24/80

Degger [83]

Answer:

1. 1/7

2. 3/10

Step-by-step explanation:

4/28 can be divided by 4

24/80 can be divided by 8

7 0

3 years ago