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Solnce55 [7]
3 years ago
8

24x + 30x = 3x + 24 + 4

Mathematics
2 answers:
chubhunter [2.5K]3 years ago
8 0

x equals about 0.549

Alinara [238K]3 years ago
7 0

24x + 30x = 3x + 24 + 4


54x=3x+24+4


54x=3x+28


54x-3x=28


51x=28


X=28/51

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Lina20 [59]
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8 0
3 years ago
Without graphing, tell whether the point (10, 37) is on the graph of y = 3x + 7.
slava [35]

Answer: YES

Plug in the numbers in your head.

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7 0
3 years ago
Help!!! Quick!!! Plzzzzz!!!
max2010maxim [7]

Answer:

x = -8

y = 9

Step-by-step explanation:

to solve this expression using simultaneous equation, we would say let

y =9............................................. equation 1

 6x + 5y =-3............................................equation 2

substitute equation 1 into equation 2

-3 = 6x + 5y............................................equation 2

6x + 5(9) = -3

6x + 45 = -3

collect the like terms

6x = -3-45

6x = -48

divide both sides by the coefficient of x which is 6

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4 0
4 years ago
Turn 10+ negative 3 3/8 into a single rational number
natta225 [31]

Answer:

the answer for this question is

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3 0
4 years ago
The mean MCAT score 29.5. Suppose that the Kaplan tutoring company obtains a sample of 40 students with a mean MCAT score of 32.
Paul [167]

Answer:

We conclude that the students that took the Kaplan tutoring have a mean score greater than 29.5.

Step-by-step explanation:

We are given that the Kaplan tutoring company obtains a sample of 40 students with a mean MCAT score of 32.2 with a standard deviation of 4.2.

Let \mu = <u><em>population mean score</em></u>

So, Null Hypothesis, H_0 : \mu \leq 29.5      {means that the students that took the Kaplan tutoring have a mean score less than or equal to 29.5}

Alternate Hypothesis, H_A : \mu > 29.5      {means that the students that took the Kaplan tutoring have a mean score greater than 29.5}

The test statistics that will be used here is <u>One-sample t-test statistics</u> because we don't know about population standard deviation;

                               T.S.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean MCAT score = 32.2

            s = sample standard deviation = 4.2

            n = sample of students = 40

So, <u><em>the test statistics</em></u> =  \frac{32.2-29.5}{\frac{4.2}{\sqrt{40} } }  ~  t_3_9

                                    =  4.066

The value of t-test statistics is 4.066.

Now, at 0.05 level of significance, the t table gives a critical value of 1.685 at 39 degrees of freedom for the right-tailed test.

Since the value of our test statistics is more than the critical value of t as 4.066 > 1.685, so <u><em>we have sufficient evidence to reject our null hypothesis</em></u> as it will fall in the rejection region.

Therefore, we conclude that the students that took the Kaplan tutoring have a mean score greater than 29.5.

3 0
3 years ago
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