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castortr0y [4]
3 years ago
12

144-120x+25x² help me​

Mathematics
1 answer:
34kurt3 years ago
7 0

Answer:

144 - 120x + 25 {x}^{2}  \\ 144 - 95 {x}^{3}

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3−2x=2x+3(6−3x)−7(1−2y)=25+6y
Zielflug [23.3K]

Answer:

Step-by-step explanation:

First you need to use the distributive property:

3-2x+2x+3(6-3x)-7(1-2y)=25+6y

3-2x=2x+18-9x-7+14y=25+6y

3-2x=-7x+11+14y=25+6y

Then Split up the equations.

3-2x=-7x+11

Subtract 3 from both sides

-2x=-7x+8

Then add -7x to both sides

5x=8

Then divide both sides by 5x

X=\frac{5}{8}

To solve for y: just use 14y=25+6y

Subtract both sides by 6y

8y=25

Then divide both sides by 8y

y=\frac{25}{8}

7 0
3 years ago
My question got deleteddd but does anyone wanna talk? I'm 14​
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7 0
3 years ago
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Can geniuses tell the right answer?​
Kazeer [188]

Answer:

under the 48 is -24

on the right side of 6 is -5

on the right side of 48 is -40

7 0
3 years ago
(6th grade math)
Montano1993 [528]

Answer:

He drank 15.93 ounces.

4 0
3 years ago
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1. The national mean (μ) IQ score from an IQ test is 100 with a standard deviation (s) of 15. The dean of a college wants to kno
Nat2105 [25]

Answer:

We conclude that the mean IQ of her students is different from the national average.

Step-by-step explanation:

We are given that the national mean (μ) IQ score from an IQ test is 100 with a standard deviation (s) of 15.

The dean of a college want to test whether the mean IQ of her students is different from the national average. For this, she administers IQ tests to her 144 students and calculates a mean score of 113

Let, Null Hypothesis, H_0 : \mu = 100 {means that the mean IQ of her students is same as of national average}

Alternate Hypothesis, H_1 : \mu\neq 100  {means that the mean IQ of her students is different from the national average}

(a) The test statistics that will be used here is One sample z-test statistics;

               T.S. = \frac{Xbar-\mu}{\frac{s}{\sqrt{n} } } ~ N(0,1)

where, Xbar = sample mean score = 113

              s = population standard deviation = 15

             n = sample of students = 144

So, test statistics = \frac{113-100}{\frac{15}{\sqrt{144} } }

                             = 10.4

Now, at 0.05 significance level, the z table gives critical value of 1.96. Since our test statistics is more than the critical value of z which means our test statistics will fall in the rejection region and we have sufficient evidence to reject our null hypothesis.

Therefore, we conclude that the mean IQ of her students is different from the national average.

3 0
4 years ago
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