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nikklg [1K]
3 years ago
5

Help plz. i would greatly appreciate it​

Mathematics
2 answers:
Likurg_2 [28]3 years ago
8 0

Answer:

1) 25 = 5²

2) x = 5²

3) 64 = b³

Step-by-step explanation:

logb(a) = x

a = b^x

log5(25) = 2

25 = 5²

log5(x) = 2

x = 5²

logb(64) = 3

64 = b³

vekshin13 years ago
6 0

Answer:

1) 5^2=25

2) 5^2=x

3) b^3=64

Step-by-step explanation:

To write logs of the form log_ba=x in their exponential form, you take the base b and put it to the power of x and then set that equal to a: b^x=a.

1. Here, b = 5, a = 25, and x = 2, so: 5^2=25

2. In this problem, b = 5, x = 2, and a = x, so: 5^2=x

3. Finally, here, b = b, a = 64, and x = 3, so: b^3=64

Hope this helps!

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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
How many square feet of carpet will we need for this hole
Inessa05 [86]

Answer:

FIND THE HOLE SIZE IN THIS CASE THERE IS NO SIZE SO  ZERO IS THE AWNSER

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3 years ago
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oee [108]

Answer:

A' 2,2

B' 3, -1

C' -1,0

Step-by-step explanation:

Because you're translating it by -2, 3, you're basically subtracting 2 from the x value and adding 3 to the y value.

A = 4, -1

A' = 2, 2

B = 5, -4

B' = 3, -1

C = 1, -3

C' = -1, 0

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15) Which statement is true?
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d

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