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melisa1 [442]
3 years ago
9

What is the value of (2.4 x 10^3)+(5.7×10^2)​

Mathematics
2 answers:
Gennadij [26K]3 years ago
8 0

Answer:0

2970

Step-by-step explanation:

nordsb [41]3 years ago
4 0

Answer:

2970

Step-by-step explanation:

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Adam is a golfer who thinks that his drives have an average length of more than 200\text{ m}200 m200, start text, space, m, end
miskamm [114]

Answer:

t=2.24 for khan

Step-by-step explanation:

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3 years ago
Teresa earned scores of 88, 84, 88 points and 91 points on 4 test what is the lowest score she can get on the fifth test and sti
Usimov [2.4K]

Answer:

99 points

Step-by-step explanation:

Average of a data a is the sum of the data divided by the number of data.

Average = sum of data ÷ number of data

A = T/n .....1

For the average of 5 test to be 90.

The total sum of score must be;

Number of tests n = 5

Average A = 90

Substituting into equation 1

90 = T/5

T = 90×5 = 450 points

The total score must be equal to 450 point to have an average of 90 points

The sum of the first four test is;

88+84+88+91 = 351

Let x represent the first score,

351 + x = 450

x = 450 - 351

x = 99 points

The minimum score she can get in the fifth test is 99 points

5 0
3 years ago
What is the domain and range of the function?<br> Domain:
ad-work [718]
It depends on the function
8 0
3 years ago
Read 2 more answers
A survey of 280 homeless persons showed that 63 were veterans. Construct a 90% confidence interval for the proportion of homeles
MA_775_DIABLO [31]

Answer:

[0.184, 0.266]

Step-by-step explanation:

Given:

Number of survey n =280

Number of veterans = 63

Confidence interval = 90%

Computation:

Probability of veterans = 63/280

Probability of veterans =0.225

a=0.1

Z(0.05) = 1.645 (from distribution table)

Confidence interval = 90%

So,

p ± Z*√[p(1-p)/n]

0.225 ± 1.645√(0.225(1-0.225)/280)

[0.184, 0.266]

4 0
3 years ago
What’s the answer to<br> cos-1(168/240)
Nezavi [6.7K]

Answer:

ans= 45.6

Step-by-step explanation:

I think the ans is 45.6

I hope it will help u...

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3 years ago
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