Sample space for the order in which 4 blocks A B C D can be lined up

What is the exponent on 10 when you right 5.3 in scientific notation?

salantis [7]

It is zero

the answer is 5.3 x 10^0
because 10^1 would force you to move the decimal place to make it 53

I hope this helps!

8 0

3 years ago

Two angles are supplementary. One angle has a measure that is five less than four times the other. What is the measure of the la

iren [92.7K]

Answer:

The larger angle is 143

Step-by-step explanation:

x + y = 180 (This is the meaning of supplementary. Two angles make up 180 degrees)

4x - 5 = y

x + (4x - 5) = 180 Remove the brackets.

x + 4x - 5 = 180

5x - 5 = 180 Add 5 to both sides

5x - 5 +5 = 180 + 5 Combine

5x = 185 Divide by 5

5x/5=185/5

x = 37

y = 4*37 - 5

y = 148 - 5

y = 143

4 0

2 years ago

= 118

oee [108]

Answer:

number of $20 bill = 4

Number of $5 bill = 5

Step-by-step explanation:

Given that:

Total number of $5 and $20 bills = 9

Let :

Number of $5 bills = x

Number of $20 bills = y

Combined worth of all bills = $105

x + y = 9 - - - (1)

5x + 20y = 105 - - - (2)

From (1)

x = 9 - y

Substitute x = 9 - y into (2)

5(9 - y) + 20y = 105

45 - 5y + 20y = 105

15y = 105 - 45

15y = 60

y = 60/15

y = 4

x = 9 - y

x = 9 - 4 = 5

Hence,

number of $20 bill = 4

Number of $5 bill = 5

6 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

Lacie is filling a bucket with water.

ella [17]

Answer: Option A) liters

Step-by-step explanation:

Liters is the standard unit of measurement for volume. Metres, on the other hand, is the standard unit of measurement for length, height, width or distance traveled.

So, since we are to me measure the amount (volume) of water in Lacie's bucket, liters is the standard unit to used.

4 0

3 years ago