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olga nikolaevna [1]
3 years ago
13

Grayson took a math quiz last week. He got 29 problems correct and 29 problems incorrect. What percentage did Grayson get correc

t?
Write your answer using a percent sign (%).
Mathematics
2 answers:
Nana76 [90]3 years ago
8 0

Answer:

50%

Step-by-step explanation:

29 right and 29 wrong. 29+29=58

Divide 29 by 58 and you get .5

Move the decimal over two to the right, and there’s the 50%.

You could also multiply .5 by 100 and you’ll still get 50!

Sphinxa [80]3 years ago
6 0

Answer:

Step-by-step explanation:

1. Find the total number of problems, incorrect and correct. 29 incorrect problems + 29 correct problems is 58 total problems.

2. Find the total number of correct problems divided by the total number of problems. 29 incorrect problems divided by 58 total problems. This is simplified to 1/2.

3. Finally, convert 1/2 into a fraction. We can do this by multiplying the fraction by 50/50 to get a fraction of 50/100, and we know that 50/100 50 percent.

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Given f(x)=x2+12x+26. Enter the quadratic function in vertex form in the box. f(x)=
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vertex(-6,-3)

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Samara has six more than three times as many pieces of candy and Sydney. Samara has 36 pieces of candy. How many does Sydney hav
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36-6/3
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Difference between a number line and a number track
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3 0
2 years ago
Consider a system with one component that is subject to failure, and suppose that we have 115 copies of the component. Suppose f
castortr0y [4]

Answer:

Step-by-step explanation:

From the given information:

the mean (\mu) = 115 \times 20

= 2300

Standard deviation = 20 \times \sqrt{115}

Standard deviation (SD) = 214.4761

TO find:

a) P(x > 3500)= P(Z > \dfrac{3500-\mu}{214.4761})

P(x > 3500)= P(Z > \dfrac{3500-2300}{214.4761})

P(x > 3500)= P(Z > \dfrac{1200}{214.4761})

P(x > 3500)= P(Z >5.595)

From the Z-table, since 5.595 is > 3.999

P(x > 3500)=1-0.9999

P(x > 3500) = 0.0001

b)

Here, the replacement time for the mean (\mu) = \dfrac{0+0.5}{2}

= 0.25

Replacement time for the Standard deviation \sigma = \dfrac{0.5-0}{\sqrt{12}}

\sigma = 0.1443

For 115 component, the mean time = (115 × 20)+(114×0.25)

= 2300 + 28.5

= 2328.5

Standard deviation = \sqrt{(115\times 20^2) +(114\times (0.1443)^2)}

= \sqrt{(115\times 400) +(114\times 0.02082249}

= \sqrt{(46000) +2.37376386}

= \sqrt{(46000) +(2.37376386)}

= \sqrt{46002.374}

= 214.482

Now; the required probability:

P(x > 4125) = P(Z > \dfrac{4125- 2328.5}{214.482})

P(x > 4125) = P(Z > \dfrac{1796.5}{214.482})

P(x > 4125) = P(Z >8.376)

P(x > 4125) =1-  P(Z

From the Z-table, since 8.376 is > 3.999

P(x > 4125) = 1 - 0.9999

P(x > 4125) = 0.0001

7 0
3 years ago
Car lot one has a car to truck ratio of 3:4. Car lot two has a car to truck ratio of 2:3. Which car lot has more cars when both
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Answer:

this means that lot 1 would have more cars. there were 3 cars in lot 1 and 2 in lot 2

Step-by-step explanation:

lot 1-          lot 2-

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if both have 12 trucks, you need to multiply the number of cars with the number you used to make the trucks 12

3:4 would turn into 9:12

2:3 would be 8:12

this means that lot 1 would have more cars. there were 3 cars in lot 1 and 2 in lot 2

4 0
3 years ago
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