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yawa3891 [41]
3 years ago
5

6 to the second power ÷ [(4.3×3)+5.1]

Mathematics
1 answer:
Black_prince [1.1K]3 years ago
8 0
2 is the answer
That’s the answer.
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The first three terms of a sequence are given. Round to the nearest thousandth (if necessary). 360, 353,346,... 360,353,346,...
Marysya12 [62]

Answer:

66

Step-by-step explanation:

the first answer is correct

3 0
2 years ago
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g The tensile 0.2 percent offset yield strength of AISI 1137 cold-drawn steel bars up to 1 inch in diameter from 2 mills and 25
Vitek1552 [10]

Answer:

107.50

Step-by-step explanation:

Given the following :

Midpoint (S)____F

S y 93 95 97 99 101 103 105 107 109 111 f 19 25 38 17 12 10 5 4 4 2

Calculating the mean and standard deviation using a calculator :

The mean(m) of the data above = 98.12

Standard deviation (sd) = 4.02

Proportion = 99% of population

From z table = 2.33

Using :

Zscore =(x - m) / sd

2.33 = (x - 98.12) / 4.02

2.33 * 4.02 = x - 98.12

9.3666 = x - 98.12

9.3666 + 98.12= x

x = 107.4866

X = 107.50

3 0
3 years ago
Engineers measure angles in gradients, which are smaller than degrees. The table shows the conversion of some angle measures in
Schach [20]

Answer:

1.11

Step-by-step explanation:

edgenuity

5 0
3 years ago
Plz answer quickly and right T^T first one gets brainliest
netineya [11]

Answer:

the 2nd answer

Step-by-step explanation:

it makes sense

4 0
3 years ago
In the past, 35% of the students at ABC University were in the Business College, 35% of the students were in the Liberal Arts Co
sukhopar [10]

Answer:

a. proportions have not changed significantly

Step-by-step explanation:

Given

Business College= 35 %

Arts College= 35 %

Education College = 30%

Calculated

Business College = 90/300= 9/30= 0.3 or 30%

Arts College= 120/300= 12/30= 2/5= 0.4 or 40%

Education College= 90/300= 9/30 = 0.3 or 30%

First we find the mean and variance of the three colleges using the formulas :

Mean = np

Standard Deviation= s= \sqrt{npq\\}

Business College

Mean = np =300*0.3= 90

Standard Deviation= s= \sqrt{npq\\}=\sqrt{0.3*0.7*300}= 7.94

Arts College

Mean = np =300*0.4= 120

Standard Deviation= s= \sqrt{npq\\}=\sqrt{0.4*0.6*300}=  8.49

Education College

Mean = np =300*0.3= 90

Standard Deviation= s= \sqrt{npq\\}=\sqrt{0.3*0.7*300}= 7.94

Now calculating the previous means with the same number of students

Business College

Mean = np =300*0.35= 105

Arts College

Mean = np =300*0.35= 105

Education College:

Mean = np =300*0.3= 90

Now formulate the null and alternative hypothesis

Business College

90≤ Mean≥105

Arts College

105 ≤ Mean≥ 120

Education College

U0 : mean= 90     U1: mean ≠ 90

From these we conclude that the  proportions have not changed significantly meaning that it falls outside the critical region.

8 0
3 years ago
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