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sammy [17]
4 years ago
8

Mrs Richards buys 8 quarts of milk in 4 days .How many gallons of milk does she buy?

Mathematics
2 answers:
Sidana [21]4 years ago
8 0
Answer :2 gallons
1 quart = 0.25 gallons
0.25 times 8 = 2
vazorg [7]4 years ago
5 0

Answer: 2 gallons

Step-by-step explanation:

1 gallon = 4 quarts

8 divided by 4= 2

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Find the answers to this activity
diamong [38]

Answer:

(224)10 = (E0)16

I dont know why it wouldnt be right this is the answer...Im acually confused now

Step-by-step explanation:

(224)10 = (E0)16

Step by step solution

Step 1: Divide (224)10 successively by 16 until the quotient is 0:

224/16 = 14, remainder is 0

14/16 = 0, remainder is 14

Step 2: Read from the bottom (MSB) to top (LSB) as E0. This is the hexadecimal equivalent of decimal number 224

5 0
3 years ago
A sphere has a volume of 36tt in3 find the radius of the sphere​
Vsevolod [243]
V=4/3 π •r^3 multiply both sides with 3
3•36π=4π • r^3
108 π=4 π • r^3

r^3 =108 π/4 π
r=3 inch
5 0
3 years ago
Read 2 more answers
Consider the set of whole numbers from 3 to 30, inclusive.
Nadusha1986 [10]
2,4,6,8,10,12,14,16,18,20,22,24,26,28,30
8 0
4 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
In 1992, jason bought a gallon of gas for $1.23. yesterday, he bought a gallon of gas for $2.19. what is the percentage increase
marishachu [46]
1992- a gallon - 1.23dollars
yesterday - 2.19 dollars

2.19dollars - 1.23 dollars = 0.96 dollars
percentage increase is (0.96 / 1.23) x 100 = 78.0%
4 0
4 years ago
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