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3 years ago

PLEASE HELP URGENT

Mathematics

1 answer:

ad-work [718]3 years ago

8 0

Pick - d. 51 minutes

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which of the following numbers is lowest or furthest to the left on the number line 48, 95, 21, 64, 53

Tamiku [17]

Answer: I am not to sure but I believe that the correct answer is 21

Step-by-step explanation: I believe this is the correct answer because on the number line the numbers generally follow a sequenced order beginning with the lowest number and increasing towards larger numbers. The lowest number in this question is 21 and in the American systems we read left to right causing the number line to begin with the lowest number from the left and going to the right. The order would be 21, 48, 53, 64, 95.

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3 years ago

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You deposit $125 in a savings account that earns 5% annual interest compounded yearly. Find the balance in the account after the

Paul [167]

125 x.5 =62.5 times 12 months equals 750

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3 years ago

Evelyn can run 3 miles in 40 minutes. Which equation can be solved to find m, the number of miles Evelyn can run in 60 minutes?

inna [77]

Answer:

m= 4.5 miles

Step-by-step explanation:

Let the number of miles Evelyn can run be m and can be modeled by the equation below

m= kx

Where x Is the Minutes

When m= 3

x= 40

3= k40

k= 3/40

If x= 60

m=3/40(60)

m = (60*3)/40

m= 180/40

m= 4.5 miles

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3 years ago

Tim's Tamales sold 2,194 tamales for $1.25 each. What were the earnings?

allochka39001 [22]

Answer:

2,742.50

Step-by-step explanation:

2194 x 1.25 = 2742.50

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3 years ago

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If you roll two fair dice repeatedly, what is the probability that you will get a sum of 4 before you get a sum of 5 ? (a) (b) (

AlexFokin [52]

Answer:

The probability that you will get a sum of 4 before you get a sum of 5 is $\frac{1}{12}$

Step-by-step explanation:

Given : If you roll two fair dice repeatedly.

To find : What is the probability that you will get a sum of 4 before you get a sum of 5 ?

Solution :

When two dice are rolled the outcomes are

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Total number of outcomes = 36

Favorable outcome get a sum of 4 before you get a sum of 5 is (1,3) ,(2,2) and (3,1) = 3

The probability that you will get a sum of 4 before you get a sum of 5 is

$P=\frac{3}{36}$

$P=\frac{1}{12}$

Therefore, The probability that you will get a sum of 4 before you get a sum of 5 is $\frac{1}{12}$

5 0

3 years ago