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Aliun [14]
3 years ago
11

A data set has a RANGE of 24 and a MEAN of 104. If the data set contains three numbers and the highest number is 118, then what

are the other two numbers in the data set?
Mathematics
1 answer:
s2008m [1.1K]3 years ago
7 0

Answer:

highest number is 118

middle number is 100

lowest number is 94

Step-by-step explanation:

R=24---between high and low is 24 numbers

M=104---average of the three numbers

highest number is 118

therefore if you subtract 24 from 118 therefore the lowest number is 94

Then i used a mean calculator and went up from 95 to 100 before the means matched.

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Assume y varies directly as x. if y=8 when x=42; find y when x=126
Soloha48 [4]
Varies directly basically means a multiple of

y = kx

where "k" is the constant of variation. You must solve for this first using the given information.

8 = 42k
8/42 = k
0.19= k

Equation: y = 0.19x
when x = 126
y = 0.19(126)
y = 24


6 0
3 years ago
Please Help!!<br> Show all work pls! <br> #15
MrRissso [65]

Step-by-step explanation:

82 + 76 + 42 = 200

42 out of 200 hmmm

200 / 2 = 100

42 / 2 = 21

21 out of 100 = 42 out of 200

=

21%

7 0
4 years ago
Read 2 more answers
Arandom sample of n1 =12 students majoring in accounting in a college of business has a mean grade-point average of 2.70 (where
antiseptic1488 [7]

Answer:

We accept the null hypothesis that the mean gpa's are equal, with the 5 percent level of significance.

Step-by-step explanation:

We have these following hypothesis:

Null

Equal means

So

\mu_{1} = \mu_{2}

Alternative

Different means

So

\mu_{1} \neq \mu_{2}

Our test statistic is:

\frac{\overline{Y_{1}} - \overline{Y_{2}}}{\sqrt{\frac{s_{1}^{2}}{N_{1}} + \frac{s_{2}^{2}}{N_{2}}}}

In which \overline{Y_{1}}, \overline{Y_{2}} are the sample means, N_{1}, N_{2} are the sample sizes and s_{1}, s_{2} are the standard deviations of the sample.

In this problem, we have that:

\overline{Y_{1}} = 2.7, s_{1} = 0.4, N_{1} = 12, \overline{Y_{2}} = 2.9, s_{2} = 0.3, N_{2} = 10

So

T = \frac{2.7 - 2.9}{\sqrt{\frac{0.4}^{2}{12} + \frac{0.3}^{2}{10}}} = -1.3383

What to do with the null hypothesis?

We will reject the null hypothesis, that is, that the means are equal, with a significante level of \alpha if

|T| > t_{1-\frac{\alpha}{2},v}

In which v is the number of degrees of freedom, given by

v = \frac{(\frac{s_{1}^{2}}{N_{1}} + \frac{s_{2}^{2}}{N_{2}})^{2}}{\frac{\frac{s_{1}^{2}}{N_{1}}}{N_{1}-1} + \frac{\frac{s_{2}^{2}}{N_{2}}}{N_{2} - 1}}

Applying the formula in this problem, we have that:

v = 20

So, applying t at the t-table at a level of 0.975, with 20 degrees of freedom, we find that

t = 2.086

We have that

|T| = 1.3383

Which is lesser than t.

So we accept the null hypothesis that the mean gpa's are equal, with the 5 percent level of significance.

7 0
3 years ago
Pls help 10 pts<br> NO LINKS
4vir4ik [10]

Answer:

58.61%

Step-by-step explanation:

Your total backyard is 18ft * 9ft = 162 ft^2

Your patio is 5ft * 8ft = 40 ft^2

Your Garden is 4ft * 7ft = 28 ft^2

Your walkway is 3ft * 9ft = 27 ft^2

Add all those areas together you have 95 ft^2

95 of 162 = 58.61%

6 0
3 years ago
Someone pls help I need this like now pls lots of points and brainliest
Serjik [45]

Answer:

  • See below

Step-by-step explanation:

<h3>#7</h3>
  • 5⁵(16²5³)³ = 5⁵(2⁴)³(5³)³ = 5⁵2¹²5⁹ = 5¹⁴2¹²
<h3>#8</h3>
  • (8⁴5³/8⁵)² = (5³/8)² = (5³)²/(2³)² = 5⁶ / 2⁶
<h3>#9</h3>
  • (5⁸3⁷/5⁴)¹⁰ = (5⁴3⁷)¹⁰ = (5⁴)¹⁰(3⁷)¹⁰ = 5⁴⁰3⁷⁰
<h3>#10</h3>

<u>Multiplying powers with same base:</u>

  • a^b*a^c=a^{b+c}
  • Same base with powers added up

<u>Multiplying powers with same exponent but different base:</u>

  • a^b*c^b=(ac)^b
  • Same exponent with bases multiplied
8 0
3 years ago
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