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2 years ago

12

Jordan's football team gained 12 yards on their first play. On the second play, the team lost 8 yards. Which expression could be

used to describe the total yards gained by Jordan's team?

2 answers:

goldenfox [79]2 years ago

4 0

The expression is 12-8. the answer is 4 yards

Alenkinab [10]2 years ago

4 0

1The expression 12 + (-8) could be used.

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Find the equation of the line that passes through points A and B.<br> y<br> X A (4, 7)<br> B (2, 3)

g100num [7]

Answer:

y = 2x - 1.

Step-by-step explanation:

The slope = (7-3)/(4-2)

= 4/2

= 2.

y - y1 = m(x - x1)

Here m = 2 , x1 = 2 and y1 = 3. So we have:

y - 3 = 2(x - 2)

y = 2x - 4 + 3

y = 2x - 1.

We have used the point (2, 3) to find the equation but we could have used (4, 7). We would have got the same answer.

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2 years ago

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Find the missing part.

sveta [45]

Answer:

The answer is $a = \frac{15}{4}$

Step-by-step explanation:

We have to:

$cos(\theta) = \frac{adjacent\ side}{hypotenuse}$

So, if we take the 60 degree angle of the smallest triangle, we have to:

$cos(60) = \frac{a}{x}$

We do not know x. But if we take the sine of the angle of 30 degrees and the main triangle we have:

$sin(30) = \frac{x}{15}\\\\x =\frac{15}{2}$

Then:

$cos(60) = \frac{a}{\frac{15}{2}}$

$a = cos(60)\frac{15}{2}$

$a = \frac{15}{4}$

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2 years ago

Harper deposited $110 into a savings account that earns 3.5% interest a year. The function (R)t models the amount of money Raymo

schepotkina [342]

Answer:

A. $2.26

Step-by-step explanation:

An equation for Harper's balance can be written similar to the one written for Raymond's balance. It will be ...

H(t) = 110(1.035)^t

For t = 2, the two balances will be ...

H(2) = 110(1.035^2) = 117.83

R(2) = 110(1.025^2) = 115.57

The difference is ...

$117.83 -115.57 = $2.26

Harper's account will have $2.26 more.

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As a quick estimate or sanity check, you can see that Harper's interest rate is 1% more than Raymond's. So, in 2 years, he will earn a little more than 2% more on his investment than Raymond earns. 2% of $110 is $2.20, so the difference can be expected to be slightly more than this.

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3 years ago

Chef Jess made a cake with an area of 93 1/2 inches if the width of the cake was 8 1/2 inches what was the length of the cake

dybincka [34]

Answer:

L= 11

Step-by-step explanation:

Formula to find the length when you know the area:

L = A/W

So the equation for this one is:

L = 93 1/2 ÷ 8 1/2

Simplify:

L = 11

To be sure of your answer you multiply the L x W = A

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Hope this helped : D

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3 years ago

You are given the following information obtained from a random sample of 5 observations. 20 18 17 22 18 At 90% confidence, you w

Margaret [11]

Answer:

a

The null hypothesis is

$H_o : \mu = 21$

The Alternative hypothesis is

$H_a : \mu< 21$

b

$\sigma_{\= x} = 0.8944$

c

$t = -2.236$

d

Yes the mean population is significantly less than 21.

Step-by-step explanation:

From the question we are given

a set of data

20 18 17 22 18

The confidence level is 90%

The sample size is n = 5

Generally the mean of the sample is mathematically evaluated as

$\= x = \frac{20 + 18 + 17 + 22 + 18}{5}$

$\= x = 19$

The standard deviation is evaluated as

$\sigma = \sqrt{ \frac{\sum (x_i - \= x)^2}{n} }$

$\sigma = \sqrt{ \frac{ ( 20- 19 )^2 + ( 18- 19 )^2 +( 17- 19 )^2 +( 22- 19 )^2 +( 18- 19 )^2 }{5} }$

$\sigma = 2$

Now the confidence level is given as 90 % hence the level of significance can be evaluated as

$\alpha = 100 - 90$

$\alpha = 10$ %

$\alpha =0.10$

Now the null hypothesis is

$H_o : \mu = 21$

the Alternative hypothesis is

$H_a : \mu< 21$

The standard error of mean is mathematically evaluated as

$\sigma_{\= x} = \frac{\sigma}{ \sqrt{n} }$

substituting values

$\sigma_{\= x} = \frac{2}{ \sqrt{5 } }$

$\sigma_{\= x} = 0.8944$

The test statistic is evaluated as

$t = \frac{\= x - \mu }{ \frac{\sigma }{\sqrt{n} } }$

substituting values

$t = \frac{ 19 - 21 }{ 0.8944 }$

$t = -2.236$

The critical value of the level of significance is obtained from the critical value table for z values as

$z_{0.10} = 1.28$

Looking at the obtained value we see that $z_{0.10}$ is greater than the test statistics value so the null hypothesis is rejected

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3 years ago