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iVinArrow [24]
3 years ago
8

Twenty seven is smaller than five times a number increased by seven

Mathematics
1 answer:
AleksandrR [38]3 years ago
5 0

Answer:

x > 4

Step-by-step explanation:

27 < 5 ( x ) + 7

27 < 5x + 7

20 < 5x

4 < x

x > 4

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Exactly how do you figure out one step equations? We're learning about these in math now, and it's very difficult for me.
Eddi Din [679]
X/4=9
*4=9*4
x=36

Hope this helps!
4 0
3 years ago
Use the relationship between the angles in the figure to answer the question.
saw5 [17]

Answer:

B.

Step-by-step explanation:

x + 90 + 61 = 180.

The m of angle vertical to x = m  < x and the 3 angles are on the same line.

6 0
3 years ago
I need help with 37, 38, and 33
Sedbober [7]
I just happened to notice that 32 is incorrect, it should be A
33.  (21x+14)/5y+4y*5y/7(3x+2)
((21x+14)*5y)/((9y*7(3x+2)
((5(21x+14)/63(3x+2)
35(3x+2)/63(3x+2)
5/9 is the answer

37. |-3--7|+|-7--3|
|-3+7|+|-7+3|
|4|+|-4|
4+4=8
8 is the correct answer

38. 99999/1 you can 99999
5 0
4 years ago
Salim had 24 homework problems.
lesantik [10]

Answer:

so im guessing  the remainder of hw problems->

Step-by-step explanation:

24-7 =17 math problems left from doing some lunch

17-7= 10 math so its 10 math problems left

6 0
3 years ago
Brainliest + 20 pts to whoever helps pls!!
Blizzard [7]

Answer:

Correct option is (a). 60.391 and 101.879; because the test statistic is in a critical region, the test rejects the null hypothesis.

Step-by-step explanation:

A Chi-square test for population variance is used to perform this test.

The standard deviation is, 2.0 minutes.

Then the variance is, 4.0 minutes.

The hypothesis for this test is:

<em>H₀</em>: The population variance of all commute times is equal to 4.0 minutes, i.e. <em>σ² </em>= 4.

<em>Hₐ</em>: The population variance of all commute times is not equal to 4.0 minutes, i.e. <em>σ² </em>≠ 4.

The test statistic is:

\chi^{2}_{calc.}=\frac{(n-1)s^{2}}{\sigma^{2}}

The critical region of this test is defined as:

Reject <em>H₀</em> if \chi^{2}_{calc.} or \chi^{2}_{calc.}>\chi^{2}_{(1-\alpha /2), (n-1)}.

The degrees of freedom is:

n-1=81-1=80

Compute the critical from a Chi-square table.

\chi^{2}_{\alpha /2, (n-1)}=\chi^{2}_{0.05, 80}=101.879\\\chi^{2}_{(1-\alpha /2), (n-1)}=\chi^{2}_{0.95, 80}=60.391\\

The test statistic value is, \chi^{2}_{calc.}=105.8.

\chi^{2}_{calc.}=105.8 > \chi^{2}_{0.05, 80}=101.879

The null hypothesis is rejected because the test statistic is in the critical region.

Thus, the correct option is (a).

6 0
3 years ago
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