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Sidana [21]
3 years ago
14

Use the drop-down menus to locate integers on the number line. A B C D 5 -4 3 2 - 1 + 3 4 1 2 5 6 7 B В C < D​

Mathematics
1 answer:
vichka [17]3 years ago
5 0

Answer:

can you take a picture of it? the format you put it in is confusing.

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Andrei [34K]

Answer:

all work is pictured and shown

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@ranga @Becki . A new car has a sticker price of $22,450, while the invoice price paid was $19,450. What is the percentage marku
densk [106]
To answer the problem given above, divide the difference of the prices by the original price and multiply the answer by 100%. This is,
                           ((22450 - 19450) / 19450) x 100% = 15.42%
Therefore, the percentage markup of the new car is approximately 15.42%.
7 0
3 years ago
What is the area of the shape?
Tresset [83]

Answer:12 units^2

Step-by-step explanation:

2*2=4

2*2*1/2=2

2*6*1/2=6

4+2+6=12

12 units^2

4 0
3 years ago
Solve for x: 4(×+10)-3=2x+13
Marat540 [252]

Answer:

Pull out like factors

  2x + 24  =   2 • (x + 12)

2x + 24 = 2x + 24

This equation is a tautology or an Identity meaning it's always true.

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
For parts a and bâ, use technology to estimate the following. âa) The critical value of t for a 90â% confidence interval with df
rjkz [21]

Answer:

a) For the 90% confidence interval the value of \alpha=1-0.9=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:

t_{\alpha/2} =\pm 2.35

b) For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:

t_{\alpha/2} =\pm 2.62

Step-by-step explanation:

Previous concepts

The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.  

The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."

Solution to the problem

Part a

For the 90% confidence interval the value of \alpha=1-0.9=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:

t_{\alpha/2} =\pm 2.35

Part b

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:

t_{\alpha/2} =\pm 2.62

3 0
4 years ago
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