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denpristay [2]
3 years ago
8

Please solve these. Will rate you 5 star! ​

Mathematics
1 answer:
ziro4ka [17]3 years ago
7 0

Answer: Hi!

To find the answer for question 24, we would first find the square root of 20 and 5. The square root of 20 is about 4.47 (I used a calculator for convenience) and the square root of 5 is about 2.24. Now we need to add these:

4.47 + 2.24 = 6.71

Now we divide 30 by this:

30/6.71 = 4.47

Now, we'll evaluate the answer choices.

a) is equal to 1.49

b) is equal to 13.4

c) is equal to 4.47

d) is equal to 26.88

The answer choice that is correct is c, because it is what we got for the problem that we were supposed to solve in the beginning.

Hope this helps!

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Simplify :<br>(3÷7x+5÷4y)2​
marissa [1.9K]

Answer:

6/7x+5/2y

Step-by-step explanation:

I actually don't know

3 0
3 years ago
the jones family paid $150 to a painting contractor to stain their 12 ft by 15 ft deck. The smith family 16 ft by 20 ft. What pr
vitfil [10]
12ft x 15ft = 180ft divided $150 = 1.2(this is how much the fts cost)
16ft x 20ft = 320ft divided by 1.2 = $266.66
8 0
3 years ago
Plz Help Brainliest To First Answer<br><br> In rhombus ABCD , m∠BCE=33∘ .<br><br> What is m∠ABE ?
Tamiku [17]

Answer:

57 deg

Step-by-step explanation:

Rhombus ABCD has diagonals AC and BD which intersect at point E.

The diagonals of a rhombus divide the rhombus into 4 congruent triangles.

If you find the measures of the angles of one of the triangles, then you know the measures of the angles of all 4 triangles.

Also, the diagonals of a rhombus are perpendicular.

Look at triangle BCE.

m<BCE = 33

m<BEC = 90

m<BCE + m<BEC + m<CBE = 180

33 + 90 + m<CBE = 180

m<CBE = 57

<ABE in triangle ABE corresponds to <CBE in triangle BCE.

m<ABE = m<CBE = 57

Answer: 57 deg

4 0
3 years ago
(4x-8)-(-3x + 10)<br> HELP
ArbitrLikvidat [17]

Answer:

7x-18

Step-by-step explanation:

brainlliest plz

7 0
3 years ago
Read 2 more answers
A highway department executive claims that the number of fatal accidents which occur in her state does not vary from month to mo
den301095 [7]

This question is incomplete, the complete question;

A highway department executive claims that the number of fatal accidents which occur in her state does not vary from month to month. The results of a study of 193 fatal accidents were recorded. Is there enough evidence to reject the highway department executive's claim about the distribution of fatal accidents between each month? Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Fatal Accidents 11 13 22 11 15 29 15 9 13 15 22 18

Step 10 of 10 : State the conclusion of the hypothesis test at the 0.1 level of significance.

Answer: H₀ : The distribution of fatal accidents between each month is same

Step-by-step explanation:

H₀ : The distribution of fatal accidents between each month is same

H₁ : The distribution of fatal accidents between each month is different

Let the los be alpha 0.10

given data ;

       Observed    Expected  

Mon   Freq (Oi)   Freq Ei       (Oi-Ei)^2 /Ei

Jan      11    16.0833        1.606649

Feb      13    16.0833        0.591105

Mar      22    16.0833        2.176598

Apr       11    16.0833        1.606649

May     15    16.0833        0.072971

Jun     29    16.0833        10.373489

Jul     15    16.0833        0.072971

Aug       9    16.0833        3.119603

Sep      13    16.0833        0.591105

Oct      15    16.0833        0.072971

Nov      22    16.0833        2.176598

Dec      18    16.0833       0.228411

Total:   193     93                22.689119

Expected Freq ⇒ 193 / 12 = 16.08333

Test Statistic, x² : 22.6891

Num Categories: 12

Degrees of freedom: 12 - 1 = 11

Critical value X² : 24.725

P-Value: 0.0195

since Chi-square value < Chi-square critical value

and P-value > alpha 0.01 so we accept H₀

Therefore we conclude that the distribution of fatal accidents between each month is same

7 0
3 years ago
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