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ivanzaharov [21]
3 years ago
11

I can’t figure out the red part please help much appreciated

Mathematics
1 answer:
Veseljchak [2.6K]3 years ago
6 0

Answer:

Step-by-step explanation:

(23,0)

23 computers

the they make 0 dollars

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Which equivalent equation is the result of combining like terms in the given equation? 2r + 55 = 6r + 7 – 2r
nikklg [1K]
I hope I help you out
6 0
3 years ago
Someone please help!! I’ll give brainliest!
snow_lady [41]

Answer:

1766.3

Step-by-step explanation:

V=4 /3πr^3

= 4/3(3.14)(7.5)^3

= 4/3(3.14)(421.88)

= 4/3(1324.69)

= 1766.25

= 1766.3

I hope this is right!

3 0
3 years ago
Given △XPS≅△DNF, find the values of x and y.
White raven [17]

Answer:

x = 3

y = 15

Step-by-step explanation:

If △XPS ≅△DNF, their corresponding sides would be congruent. This implies that:

XP ≅ DN

PS ≅ NF

XS ≅ DF

Given that:

XP = 4y - 3

DN = 57

NF = 51

XS = 17x + 3

DF = 54

Therefore:

XP = DN

4y - 3 = 57 (Substitution)

Add 3 to both sides

4y = 57 + 3

4y = 60

Divide both sides by 4

y = 60/4

y = 15

Also,

XS = DF

17x + 3 = 54 (substitution)

Subtract 3 from each side

17x = 54 - 3

17x = 51

Divide both sides by 17

x = 51/17

x = 3

4 0
3 years ago
Find the area of each figure. Round your answers to the nearest hundredth 7-10​
stira [4]

Answer:

7. 648

8. 759.88

9. 193.52

10. 148.92

5 0
3 years ago
Read 2 more answers
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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