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3 years ago

1. 12 pizzas cost $20. What will 20 pizzas costa

Mathematics

2 answers:

Brut [27]3 years ago

4 0

Answer:

$33.33 (rounded to nearest cent)

Step-by-step explanation:

12 pizzas --- $20

20 pizzas --- $\frac{20}{12}$ × $20 = $33.33 (rounded to nearest cent)

SpyIntel [72]3 years ago

3 0

Answer:

$33.20

Step-by-step explanation:

1. Calculate the price of one pizza by dividing the cost by the amount of pizzas bought. This gives you the unit price.

$20 / 12 = $1.66 per pizza

2. Multiply the unit price by the new amount of pizzas.

$1.66 per pizza x 20 pizzas = $33.20

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-x/6 + 6 = 32 What is the answer and how do I solve it

masha68 [24]

Answer:

x= -120

Step-by-step explanation:

-x/6+6+6=32

Use -a/b=a/-b= -a/-b to rewrite the fraction.
-x/-6+6+6=32
Add the two 6's then you get 12
-x/-6+12=32
Multiply both sides with the equation by 6
-x+72=192
Move the constant to the right-hand side and change its side
-x=192+72
Subtract the numbers, -x=192-72
-x=120
Change the signs on both sides of the equation
You have your answer it is x= -120

8 0

3 years ago

Determine the t critical value(s) that will capture the desired t-curve area in each of the following cases: a. Central area 5 .

Flauer [41]

Answer:

a) "=T.INV(0.025,10)" and "=T.INV(1-0.025,10)"

And we got $t_{\alpha/2}=-2.228 , t_{1-\alpha/2}=2.228$

b) "=T.INV(0.025,20)" and "=T.INV(1-0.025,20)"

And we got $t_{\alpha/2}=-2.086 , t_{1-\alpha/2}=2.086$

c) "=T.INV(0.005,20)" and "=T.INV(1-0.005,20)"

And we got $t_{\alpha/2}=-2.845 , t_{1-\alpha/2}=2.845$

d) "=T.INV(0.005,50)" and "=T.INV(1-0.005,50)"

And we got $t_{\alpha/2}=-2.678 , t_{1-\alpha/2}=2.678$

e) "=T.INV(1-0.01,25)"

And we got $t_{\alpha}= 2.485$

f) "=T.INV(0.025,5)"

And we got $t_{\alpha}= -2.571$

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

We will use excel in order to find the critical values for this case

Determine the t critical value(s) that will capture the desired t-curve area in each of the following cases:

a. Central area =.95, df = 10

For this case we want 0.95 of the are in the middle so then we have 1-0.95 = 0.05 of the area on the tails. And on each tail we will have $\alpha/2=0.025$ .

We can use the following excel codes:

"=T.INV(0.025,10)" and "=T.INV(1-0.025,10)"

And we got $t_{\alpha/2}=-2.228 , t_{1-\alpha/2}=2.228$

b. Central area =.95, df = 20

For this case we want 0.95 of the are in the middle so then we have 1-0.95 = 0.05 of the area on the tails. And on each tail we will have $\alpha/2=0.025$ .

We can use the following excel codes:

"=T.INV(0.025,20)" and "=T.INV(1-0.025,20)"

And we got $t_{\alpha/2}=-2.086 , t_{1-\alpha/2}=2.086$

c. Central area =.99, df = 20

For this case we want 0.99 of the are in the middle so then we have 1-0.99 = 0.01 of the area on the tails. And on each tail we will have $\alpha/2=0.005$ .

We can use the following excel codes:

"=T.INV(0.005,20)" and "=T.INV(1-0.005,20)"

And we got $t_{\alpha/2}=-2.845 , t_{1-\alpha/2}=2.845$

d. Central area =.99, df = 50

For this case we want 0.99 of the are in the middle so then we have 1-0.99 = 0.01 of the area on the tails. And on each tail we will have $\alpha/2=0.005$ .

We can use the following excel codes:

"=T.INV(0.005,50)" and "=T.INV(1-0.005,50)"

And we got $t_{\alpha/2}=-2.678 , t_{1-\alpha/2}=2.678$

e. Upper-tail area =.01, df = 25

For this case we need on the right tail 0.01 of the area and on the left tail we will have 1-0.01 = 0.99 , that means $\alpha =0.01$

We can use the following excel code:

"=T.INV(1-0.01,25)"

And we got $t_{\alpha}= 2.485$

f. Lower-tail area =.025, df = 5

For this case we need on the left tail 0.025 of the area and on the right tail we will have 1-0.025 = 0.975 , that means $\alpha =0.025$

We can use the following excel code:

"=T.INV(0.025,5)"

And we got $t_{\alpha}= -2.571$

8 0

3 years ago

If anyone needs something to talk about or just needs somebody to talk too, You can comment and I'll reply as soon as possible.

vladimir2022 [97]

Answer:

okurrrrr

Step-by-step explanation:

6 0

3 years ago

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HELP!!!!!!! What is 2 8/10 + (-3.5) in decimal form

GarryVolchara [31]

Answer: -0.7

Step-by-step explanation:

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3 years ago

If sin x = .21, what is cos x

nika2105 [10]

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