Pls help will give brainlist

-15

Sloan [31]

Try the first option if that's an option

3 0

3 years ago

Mike purchased 1.35 pounds of steak for $13.50.

AlladinOne [14]

Answer:

y=10x

Step-by-step explanation:

it costs 10 dollars for one pound of steak.

3 0

3 years ago

Read 2 more answers

A certain forest covers an area of 4800 km^2 . Suppose that each year this area decreases by 5.25% . What will the area be after

9966 [12]

Answer:

the area after 6 years is 3,473 km^2

Step-by-step explanation:

The computation of the area after 6 years is as follows:

= Area × (1 - decreased percentage)^number of years

= 4,800 km^2 × (1 - 5.25%)^6

= 4,800 km^2 × 0.9475^6

= 3,473 km^2

Hence, the area after 6 years is 3,473 km^2

5 0

3 years ago

How do you calculate the midpoint of question e)? I am very desperate. please help

andrey2020 [161]

Answer:

(-1,-1)

Step-by-step explanation:

Midpoint equation: $(\frac{x_2+x_1}{2}, \frac{y_2+y_1}{2})$

= $(\frac{-5/2+1/2}{2},\frac{-1/2-3/2}{2} )$

= $(\frac{-2}{2},\frac{-2}{2})$

=(-1,-1)

Use the formula and plug it in :)

7 0

3 years ago

The degenerative disease osteoarthritis most frequently affects weight-bearing joints such as the knee. Two random samples from

Vinvika [58]

Answer:

$t=\frac{(810-770)-0}{\sqrt{\frac{67^2}{13}+\frac{56^2}{19}}}}=1.771$

$df=n_1 +n_2 -2=13+19-2=30$

Since is a right tailed test the p value would be:

$p_v =P(t_{30}>1.771)=0.0434$

Comparing the p value with the significance level $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for group 1 is significantly higher than the mean for the group 2

Step-by-step explanation:

Data given

$\bar X_{1}=810$ represent the mean for sample 1

$\bar X_{2}=770$ represent the mean for sample 2

$s_{1}=67$ represent the sample standard deviation for 1

$s_{2}=56$ represent the sample standard deviation for 2

$n_{1}=13$ sample size for the group 2

$n_{2}=19$ sample size for the group 2

$\alpha=0.05$ Significance level provided

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean for category 1 is higher than the mean for category 2, the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{2}\leq 0$

Alternative hypothesis: $\mu_{1} - \mu_{2}> 0$

We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=13+19-2=30$

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

With the info given we can replace in formula (1) like this:

$t=\frac{(810-770)-0}{\sqrt{\frac{67^2}{13}+\frac{56^2}{19}}}}=1.771$

P value

Since is a right tailed test the p value would be:

$p_v =P(t_{30}>1.771)=0.0434$

Comparing the p value with the significance level $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for group 1 is significantly higher than the mean for the group 2

4 0

3 years ago