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jolli1 [7]
3 years ago
10

Number 2 rewrite as a single power

Mathematics
1 answer:
Paul [167]3 years ago
4 0

Answer:

-625

Step-by-step explanation:

I think so..

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Write these numbers in standard notation. 3.05 x 10–3 8.92 x 106
Bad White [126]

3.05  \times   {10}^{ - 3} =0.00305 \\8.92 \times  {10}^{6} =   8920000
3 0
3 years ago
Read 2 more answers
PLEASE help I’ll mark you as the Brainiest please!!!!
puteri [66]

Answer:

c

Step-by-step explanation:

6 0
3 years ago
Oliver interviewed 30% of the 9th grade class and 70% of the 10th grade class at his school. Jenny interviewed 75% of the 9th gr
julia-pushkina [17]

Answer:

A. 36

Step-by-step explanation:

We are given a total of 176 interviewed by Oliver and a total of 140 interviewed by Jenny. To find how many more 10th graders than 9th graders were interviewed, subtract the totals given

176 - 140 = 36

This is how we came to the answer:

We are given 70% of the 10th-grade and 30% of the 9th-grade with a total of 176 for Oliver.

While we're given 75% of the 9th-grade class and 25% of the 10th-grade with a total of 140 interviewed by Jenny

Oliver's Interviewees

10-graders

Firstly, let's find what the number of 9th-graders was interviewed by Oliver; find the percentage of the 9th-graders by the total;

70% of 176 =

Cross multiply

123.2 were 10-graders interviewed by Oliver

9th-graders

Now, to find the number of 9th-graders was interviewed by Oliver; find the percentage of the 9th-graders by the total;

30% of 176 =

 

Cross multiply

52.8 were 9th-graders interviewed by Oliver

Jenny's Interviewees

9th-graders

Firstly, let's find what the number of 9th-graders was interviewed by Jenney; find the percentage of the 9th-graders by the total;

75% of 140 =

 

Cross multiply

105 students were 9th-graders interviewed by Jenney.

10th-graders

Now, to find the number of 10th-graders was interviewed by Jenney; find the percentage of the 10th-graders by the total;

25% of 140 =  

 

Cross multiply  

35 students were 10th-graders interviewed by Jenney.  

Total calculation

Use the results and sum them up by 9th-grade plus 9th-grade and 10th-grade plus 10-grade. Then subtract the amount gotten from 9th-grade away from the amount gotten from 10th-grade;

Oliver's 9th-grade = 52.8

Jenny's 9th-grade = 105

105 + 52.8 = 157.8

Oliver's 10th-grade = 123.2

Jenny's 10th-grade = 35

123.2 + 35 = 158.2

Total calculation: 158. 2 - 157.8 = 0.4

Therefore, there are 36 more 10th than 9th.

For more information, visit: brainly.com/question/23490909

7 0
2 years ago
I need help on questions 48 and 49.
xenn [34]
48. A square's sides are equal, so we know that its area is a number multiplied by itself, or squared. To find the length of one side, you must take the square root of its area, 156.25, which is 12.5. To find the square's perimeter (how much fence we need to go around the garden) you must multiply that side by 4, to get 50. Antonine has 40 ft already, so he only needs 10 more ft, and each ft costs 4.97. Our final answer is 10*4.97, or $49.70. 

49. The square root of 12 is about 3.46, plus 2 is our answer, about 5.46 cm.
8 0
3 years ago
Independent random samples of vehicles traveling past a given point on an interstate highway have been observed on monday versus
quester [9]
Hi! 

To compare this two sets of data, you need to use a t-student test:

You have the following data:

-Monday n1=16; <span>x̄1=59,4 mph; s1=3,7 mph

-Wednesday n2=20;  </span>x̄2=56,3 mph; s2=4,4 mph

You need to calculate the statistical t, and compare it with the value from tables. If the value you obtained is bigger than the tabulated one, there is a statistically significant difference between the two samples.

t= \frac{X1-X2}{ \sqrt{ \frac{(n1-1)* s1^{2}+(n2-1)* s2^{2} }{n1+n2-2}} * \sqrt{ \frac{1}{n1}+ \frac{1}{n2}} } =2,2510

To calculate the degrees of freedom you need to use the following equation:

df= \frac{ (\frac{ s1^{2}}{n1} + \frac{ s2^{2}}{n2})^{2}}{ \frac{(s1^{2}/n1)^{2}}{n1-1}+ \frac{(s2^{2}/n2)^{2}}{n2-1}}=33,89≈34

The tabulated value at 0,05 level (using two-tails, as the distribution is normal) is 2,03. https://www.danielsoper.com/statcalc/calculator.aspx?id=10

So, as the calculated value is higher than the critical tabulated one, we can conclude that the average speed for all vehicles was higher on Monday than on Wednesday.



5 0
3 years ago
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