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Gnesinka [82]
2 years ago
9

5x+2y=-3

Mathematics
1 answer:
max2010maxim [7]2 years ago
7 0
The main idea is let y be a subject and bring in the first equation and now it just have one letter in first equation, when you resolve the value of x, finally put the value of x into first or second equation then you can calculate what the value of y

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Númer 9 and 10 please help
vredina [299]
9. -1,07, 1/2. 3........it is all I could see

10. -V25= -5
-5, -4,3, 0, V15, 4,2
8 0
3 years ago
An art history professor assigns letter grades on a test according to the following scheme. A: Top 13%13% of scores B: Scores be
amm1812

Answer:

The numerical limits for a B grade is between 81 and 89.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 79.7, \sigma = 8.4

B: Scores below the top 13% and above the bottom 56%

Below the top 13%:

Below the 100-13 = 87th percentile. So below the value of X when Z has a pvalue of 0.87. So below X when Z = 1.127. So

Z = \frac{X - \mu}{\sigma}

1.127 = \frac{X - 79.7}{8.4}

X - 79.7 = 8.4*1.127

X = 89

Above the bottom 56:

Above the 56th percentile, so above the value of X when Z has a pvalue of 0.56. So above X when Z = 0.15. So

Z = \frac{X - \mu}{\sigma}

0.15 = \frac{X - 79.7}{8.4}

X - 79.7 = 8.4*0.15

X = 81

The numerical limits for a B grade is between 81 and 89.

3 0
3 years ago
Andrew has 212 coins, all of
Igoryamba

Answer:

158 quarters

54 dimes

Step-by-step explanation:

Let d = no. of dimes

     q = no. of quarters

10d = value of all the dimes since each dime is worth 10 cents

and 25q = value of all the quarters since each quarter is worth 25 cents

$44.90 = 4490 cents

  10d + 25q = 4490              d + q = 212

<u> - 10d - 10q = -2120</u>

             15q = 2370

                 q = 158 quarters   d + 158 = 212

                                                        d = 54 dimes

8 0
3 years ago
1.What is the intersection of all the open intervals containing the closed interval [0,1]? Justify your answer.
ExtremeBDS [4]

Answer:

Step-by-step explanation:

1)

We can find an open interval (-\frac{1}{n}, 1+\frac{1}{n}). Which contains the closed interval [0,1] when n \to \infinity

Therefore, the intersection of all the open intervals containing [0,1[/tex[ is [tex][0,1]

2)

We can find closed intervals containing [0,1] in (-\frac{1}{n}, 1+\frac{1}{n}) when n \to \infinity

Since (-\frac{1}{n}, 1+\frac{1}{n}) is closed, the intersection of all those intervals containing [0,1[/tex[ is  the closed interval [tex][0,1]

7 0
3 years ago
2(x + 5) = 3x + 1<br> Solve the equation <br><br> Please hurry
Novosadov [1.4K]

Answer:

So the answer is 9

Step-by-step explanation

1 . The explanation is that you have to distrubte the parentheses 2.subtract 10 from both sides 3.simplify what you got after you 4.subtract 3 from both sides 5.simplify again 6.divide both sides of the equation by the same terms 7.simplify again and you get 9

8 0
3 years ago
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