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2 years ago

12

Barbara has $70.00 in her pocket and a $20 gift card. which purchase will she be able to make without having to use her credit c

ard?

2 answers:

Tresset [83]2 years ago

4 0

Answer:

Step-by-step explanation:

six puzzles for $11.99 each and a board game for $15.89

Kipish [7]2 years ago

3 0

If Barbara has $ 70.00 in her pocket and a $ 20 gift card. then the maximumm amount she can pruchase is $ 70 + $ 20 = $ 90
so let x be purchase will she be able to make without having to use her credit card
x >= $ 90

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TJ invested $4000 in a bond at a yearly rate of 2%. He earned $200 in interest. How long was the money invested

kow [346]

4000.00 x 2.0 = $80.00 per year x 2 years = $160.00 so 6 months would earn $40.00 taking it to the $200, so 2 yrs 6 mos.

3 0

3 years ago

Twenty five heat lamps are connected in a greenhouse so, that when one lamp fails, another takes over immediately (only one lamp

olga55 [171]

Answer:

P = 0.006

Step-by-step explanation:

Given

n = 25 Lamps

each with mean lifetime of 50 hours and standard deviation (SD) of 4 hours

Find probability that the lamp will be burning at end of 1300 hours period.

As we are not given that exact lamp, it means we have to find the probability where any of the lamp burning at the end of 1300 hours, So we have

Suppose i represents lamps

P (∑i from 1 to 25 ( $X_i$ > 1300)) = 1300

= P( $X^{'}$ > $\frac{1300}{25}$ ) where $X^{'}$ represents mean time of a single lamp

= P (Z> $\frac{52-50}{\frac{1}{\sqrt{25}}}$ ) Z is the standard normal distribution which can be found by using the formula

Z = Mean Time ( $X^{'}$ ) - Life time of each Lamp (50 hours)/ (SD/ $\sqrt{n}$ )

Z = (52-50)/(4/ $\sqrt{25}$ ) = 2.5

Now, P(Z>2.5) = 0.006 using the standard normal distribution table

Probability that a lamp will be burning at the end of 1300 hours period is 0.006

4 0

3 years ago

I am making hamburgers that are 3/4 pounds each. If I have 30 pounds of Hamburg, what fraction of hamburger do I have left.

erma4kov [3.2K]

Answer:

1/4

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

#12 with explanation please

4vir4ik [10]

Answer:

30.01 units

Step-by-step explanation:

<u>Given vertices WXYZ:</u>

W(-4,5), X(2,4), Y(3, -4), Z(-1, -6)

Perimeter is the sum of the side length.

<u>Use distance formula to find each side:</u>

WX = $\sqrt{(2+4)^2+(4-5)^2} =\sqrt{37}$ ≈ 6.08
XY = $\sqrt{(3-2)^2+(-4-4)^2} =\sqrt{65}$ ≈ 8.06
YZ = $\sqrt{(-1-3)^2+(-6+4)^2} =\sqrt{20}$ ≈ 4.47
WZ = $\sqrt{(-1+4)^2+(-6-5)^2} =\sqrt{130}$ $\sqrt{(-1+4)^2+(-6-5)^2}=\sqrt{130}$ ≈ 11.40

<u>Find perimeter:</u>

P = 6.08 + 8.06 + 4.47 + 11.40 = 30.01 units

7 0

2 years ago

student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$

= $(\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5$

= $(\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]$

= $(\frac{3}{4^4})[1+\frac{1}{16}]$

= $(\frac{3}{256})[\frac{17}{16}]$

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0

2 years ago