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3 years ago

7

1/4 as a fraction and a decimal

1 answer:

storchak [24]3 years ago

8 0

Answer: 1/4 fraction .25 decimal

Hope this helps :)

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The sum of the digits of a two-digit number is 8. The difference between the number and the reversed number is 10 more than the

Vesna [10]

Answer:

The number is 62

Step-by-step explanation:

Let the digits of the number be T and U;

A 2 digit number is represented as: 10T + U

So,

$T + U = 8$

$10T + U - (10U + T) = 10 + 10U + T$

Required

Find the digit

Make U the subject in the first equation

$U = 8 - T$

Substitute 8 - T for U in the second

$10T + U - (10U + T) = 10 + 10U + T$

$10T + 8 - T - (10*(8-T) + T) = 10 + 10(8 - T) + T$

$10T + 8 - T - (80-10T + T) = 10 + 80 - 10T+ T$

$10T + 8 - T -80+10T - T = 10 + 80 - 10T+ T$

Collect Like Terms

$10T + 10T - T- T + 8 -80 = 10 + 80 - 10T+ T$

$18T -72 = 90 - 9T$

$9T + 18T = 90 + 72$

$27T = 162$

$T = 162/27$

$T = 6$

Recall that:

$U = 8 - T$

$U = 8 - 6$

$U = 2$

<em>Hence, the number is 62</em>

5 0

3 years ago

The probability that a randomly selected 2 2-year-old male garter snake garter snake will live to be 3 3 years old is 0.98861 0

Mnenie [13.5K]

Answer:

a. $Probability = 0.97735$

b. $Probability = 0.92294$

c. $P(At\ Least\ One) = 1$

No, it is not unusual if at least 1 lives up to 3.

Step-by-step explanation:

Given

Represent the probability that a 2 year old snake will live to 3 with P(Live);

$P(Live) = 0.98861$

Solving (a): Probability that two selected will live to 3 years.

Both snakes have a chance of 0.98861 to live up to 3 years.

So, the required probability is:

$Probability = P(Live)\ and\ P(Live)$

$Probability = 0.98861 * 0.98861$

$Probability = 0.9773497321$

$Probability = 0.97735$ <em>--- Approximated</em>

Solving (b): Probability that seven selected will live to 3 years.

All 7 snakes have a chance of 0.98861 to live up to 3 years.

So, the required probability is:

$Probability = P(Live)^n$

Where $n = 7$

$Probability = 0.98861^7$

$Probability = 0.92294324145$

$Probability = 0.92294$ <em>--- Approximated</em>

Solving (c): Probability that at least one of seven selected will not live to 3 years.

In probabilities, the following relationship exist:

$P(At\ Least\ One) = 1 - P(None).$

So, first we need to calculate the probability that none of the 7 lived up to 3.

If the probability that one lived up to 3 years is 0.98861, then the probability than one do not live up to 3 years is 1 - 0.98861

This gives:

$P(Not\ Live) = 0.01139$

The probability that none of the 7 lives up to 3 is:

$P(None) = P(Not\ Live)^7$

$P(None) = 0.01139^7$

Substitute this value for P(None) in

$P(At\ Least\ One) = 1 - P(None).$

$P(At\ Least\ One) = 1 - 0.01139^7$

$P(At\ Least\ One) = 0.99999999999997513055642436060443621$

$P(At\ Least\ One) = 1$ ---- Approximated

No, it is not unusual if at least 1 lives up to 3.

This is so because the above results, which is 1 shows that it is very likely for at least one of the seven to live up to 3 years

7 0

3 years ago

Preform the indicated operation.

Tems11 [23]

Answer:

34325

Step-by-step explanation:

3 0

3 years ago

What is y= X/2 + 9; x= -12

Lelu [443]

-12/2=-6
y=-6+9
-6+9=3
y=3

4 0

3 years ago

Amy and John shared 150 peaches. (1/3) of what Amy has equals to (1/2) of what John has. How many does Amy have? How many does J

frosja888 [35]

If my math is correct amy has 90 and john has 60

8 0

4 years ago

Read 2 more answers