Mark simplified three over five ÷ five over eight; his work is shown below. Identify where he made his error. (5 points)

What is the net force on the image(NEED ASAP)

ollegr [7]

Answer:

1200 N right

Step-by-step explanation:

Subtract 150 from 1350. Since there is more force towards the right, the net force is 1200 N right

8 0

3 years ago

Read 2 more answers

Which linear function is represented by the graph?

Vesnalui [34]

Answer:

f(x)could also be written as y in y= mx+ c

by taking two points we are gonna calculate.

(0,1), (4,-1)

finding m

y*- y'/x*-x'

=1-(-1)/0-4

=2/-4

=-2

now c ( intercept)

take any point

eg. (0,1)

y=mx+c

1=-2×0+c

1=c

equation then equals to

y=-2x+1

f (x) =-2x+1

4 0

3 years ago

Plz help!!!! #4 I can’t figure this out it’s the picture!

eimsori [14]

Answer:

I think the answer is that Best Market sells strawberries at a higher unit price, and that you would have to pay $5.40 more at that store than the other.

Step-by-step explanation:

If we take a look at the graph, we can see that 1 pound of strawberries is worth 2$. That makes the unit price, or the price of per one unit of the item, 2$ for Fine Foods.

In thd directions, they gave us the equation y = 2.9x for strawberries at Best Market. To find the unit price for Best Market, we would have to substitute x for 1 since x represent pounds, and we're trying to find how much 1 pound costs to find the unit price. It would look like this:

y = 2.9 (1)

y = 2.9

So, the unit price at Best Market would be 2.90$.

Comparing the unit prices, 2.90$ is bigger than &2. So, Best Market sells strawberries at a higher unit price. To find out how much 6 pounds of strawberries at Best Market is, all we have to do is substitute x for 6 in y = 2.9x like so:

y = 2.9 (6)

y = 17.40

So, the price for 6 pounds of strawberries is 17.40$ at Best Market. If we refer to the graph for the cost of 6 pounds at Fine Foods, then we know that 6 pounds of strawberries is 12$. 17.40 - 12 = 5.4. So, you would have to pay $5.40 more for 6 pounds at Best Market than Fine Foods.

3 0

3 years ago

A certain test preparation course is designed to help students improve their scores on the LSAT exam. A mock exam is given at th

ASHA 777 [7]

Answer:

The critical value is $T_c = 2.5706$ .

The 90% confidence interval for the average net change in a student's score after completing the course is (9.04, 18.30).

Step-by-step explanation:

Before building the confidence interval, we need to find the sample mean and the sample standard deviation:

Sample mean:

$\overline{x} = \frac{6+16+19+12+15+14}{6} = 13.67$

Sample standard deviation:

$s = \sqrt{\frac{(6-13.67)^2+(16-13.67)^2+(19-13.67)^2+(12-13.67)^2+(15-13.67)^2+(14-13.67)^2}{5}} = 4.4121$

Confidence interval:

We have the standard deviation for the sample, so the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 6 - 1 = 5

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 5 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.9}{2} = 0.95$ . So we have T = 2.5706, that is, the critical value is $T_c = 2.5706$

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.5706\frac{4.4121}{\sqrt{6}} = 4.63$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 13.67 - 4.63 = 9.04.

The upper end of the interval is the sample mean added to M. So it is 13.67 + 4.63 = 18.30.

The 90% confidence interval for the average net change in a student's score after completing the course is (9.04, 18.30).

7 0

3 years ago

What is 8n=3m 1;n=-2,2,4

algol [13]

8n = 3m + 1
~~~~~~~~~~~~~~~~~~~
8(-2) = 3m + 1
-16 = 3m + 1
-1 -1
-17 = 3m
--- ---
3 3
-17/3 = m
~~~~~~~~~~~~~~~~~~~
8(2) = 3m + 1
16 = 3m + 1
-1 -1
15 = 3m
--- ---
3 3
5 = m
~~~~~~~~~~~~~~~~~~~
8(4) = 3m + 1
32 = 3m + 1
-1 -1
31 = 3m
--- ---
3 3
31/3 = m

5 0

3 years ago