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Lesechka [4]
3 years ago
11

Wyatt is training for a race. He will run 3/4 of a mile a day for 8 days. How far will Wyatt run altogether?

Mathematics
1 answer:
Natasha2012 [34]3 years ago
8 0

Answer:

6 miles

Step-by-step explanation:

3/4 x 8

=6

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Expand the following: 3(16x-24)
Fynjy0 [20]

Answer:

48x-72

Step-by-step explanation:

u multiply 3 by 16x then multiply 3 by 24 and bring down your  minus sign

6 0
3 years ago
Reuben​ says, "two days ago i was 20 years old. later next​ year, i will be 23 years​ old." how is this​ possible?
Aneli [31]
Let us always base on the present age. We denote this by x. Now, the age he had 2 years ago would then be denoted as (x-2). Let's equate this to 20 years old.

x - 2 = 20
x = 20 + 2
x = 22 years old
He should be 22 years old now.

Let's check the other condition. After 1 year, his age should be (x+1). Let's equate this to 23 years old.

x + 1 = 23
x = 23 - 1
x = 22


Thus, this is possible if Reuben is 22 years old as of the present.
7 0
3 years ago
How to solve this I am not the best at math
Brrunno [24]
Try  it is a great site that you can plug problems in to for the answers! I hope this is helpful!!
5 0
3 years ago
2/5 + r = 1 and 4/5 <br> What is r?
MariettaO [177]
Answer- r= 7\5 hope this helps!!
7 0
3 years ago
During a recent drought, a water utility in a certain town sampled 100 residential water bills and found out that 73 of the resi
ankoles [38]

Answer:

(0.6430, 0.8170)

Step-by-step explanation:

Given that during a recent drought a water utility in a certain town sampled 100 residential water bills and found out that 73 of the residences had reduced their water consumption over that of the previous year.

Sample size n = 100

Sample proportion p = \frac{73}{100} =0.73

q = 1-p = 0.23

Std error of proportion = \sqrt{\frac{pq}{n} } \\=\sqrt{\frac{0.73*0.27}{100} } \\=0.04440

95% Z critical value = 1.96

Margin of error = 1.96*0.0444 = 0.0870

Confidence interval = sample proportion ±margin of error

0.642983946

0.817016054

(0.6430, 0.8170)

4 0
2 years ago
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