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Musya8 [376]
3 years ago
6

The midpoint of line MN is P(5,5). One of the end points in M(-4,2). Find the coordinate of endpoint N.

Mathematics
1 answer:
Greeley [361]3 years ago
7 0
-4 + x                        2 + y
-------  =  5  and         ------  = 5
   2                               2

so x = 10 + 4 = 14  and y = 10 - 2 = 8

endpoint N = (14,8)
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Which number, when rounded to the nearest tenth, is 43.6? Options: A: 43.55 B: 43.657 C: 43.65 D: 43.549
weqwewe [10]

Answer:

A: 43.55

Step-by-step explanation:

Place value can be defined as the numerical value representing a digit owing to its position in a number. Some examples of the various place values used in representing each digit are tenths, hundredths, thousandths, unit, tens, hundreds, thousands etc.

For example, in the number 43.55;

  • The place value of 4 is tens.
  • The place value of 3 is units.
  • The place value of 5 is tenths.
  • The place value of 5 is hundredths.

Rounding up to the nearest tenth simply means to have 1 significant figure after the decimal point.

Therefore, 43.55 to the nearest tenth is 43.6

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3 years ago
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Angelina_Jolie [31]
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Evaluate 3(x-1)^2 if x=4
Ronch [10]

Answer:

27

Step-by-step explanation:

3(4-1)²

3(3)²

3 x 9

= 27

5 0
3 years ago
Which of the following shows how to write 8,240 pennies as<br> dollars?
olasank [31]

Answer:

$ 82.40

Step-by-step explanation:

you divide

100 pennies in a dollar

8240/100=82.40

5 0
2 years ago
Here are summary statistics for randomly selected weights of newborn​ girls: nequals202​, x overbarequals28.3 ​hg, sequals6.1 hg
Lorico [155]

Answer:

The confidence interval is 27.5 hg less than mu less than 29.1 hg

(A) Yes, because the confidence interval limits are not similar.

Step-by-step explanation:

Confidence interval is given as mean +/- margin of error (E)

mean = 28.3 hg

sd = 6.1 hg

n = 202

degree of freedom = n-1 = 202-1 = 201

confidence level (C) = 95% = 0.95

significance level = 1 - C = 1 - 0.95 = 0.05 = 5%

critical value corresponding to 201 degrees of freedom and 5% significance level is 1.97196

E = t×sd/√n = 1.97196×6.1/√202 = 0.8 hg

Lower limit = mean - E = 28.3 0.8 = 27.5 hg

Upper limit = mean + E = 28.3 + 0.8 = 29.1 hg

95% confidence interval is (27.5, 29.1)

When mean is 28.3, sd = 6.1 and n = 202, the confidence limits are 27.5 and 29.1 which is different from 27.8 and 29.6 which are the confidence limits when mean is 28.7, sd = 1.8 and n = 17

7 0
3 years ago
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