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gulaghasi [49]
3 years ago
6

point K is between points M and Q. If line segment MK = 7 inches and line segment MQ = 15 inches, whst is the measurement of KQ?

Mathematics
1 answer:
BigorU [14]3 years ago
8 0

The total length of a line segment is the sum of the lengths of its parts.

MQ = MK + KQ . . . . . . express the relationship between the segments

15 in = 7 in + KQ . . . . . fill in the given information

(15 - 7) in = KQ . . . . . . .subtract 7 in

KQ = 8 in . . . . . . . . . . . simplify

The measurement of KQ is 8 inches.

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A random sample of 10 parking meters in a resort community showed the following incomes for a day. Assume the incomes are normal
GenaCL600 [577]

Answer:

A 95% confidence interval for the true mean is [$3.39, $6.01].

Step-by-step explanation:

We are given that a random sample of 10 parking meters in a resort community showed the following incomes for a day;

Incomes (X): $3.60, $4.50, $2.80, $6.30, $2.60, $5.20, $6.75, $4.25, $8.00, $3.00.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                         P.Q.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean income = \frac{\sum X}{n} = $4.70

            s = sample standard deviation = \sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }  = $1.83

            n = sample of parking meters = 10

            \mu = population mean

<em>Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.</em>

<u>So, 95% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-2.262 < t_9 < 2.262) = 0.95  {As the critical value of t at 9 degrees of

                                            freedom are -2.262 & 2.262 with P = 2.5%}  

P(-2.262 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 2.262) = 0.95

P( -2.262 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu < 2.262 \times {\frac{s}{\sqrt{n} } } ) = 0.95

P( \bar X-2.262 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+2.262 \times {\frac{s}{\sqrt{n} } } ) = 0.95

<u>95% confidence interval for</u> \mu = [ \bar X-2.262 \times {\frac{s}{\sqrt{n} } } , \bar X+2.262 \times {\frac{s}{\sqrt{n} } } ]

                                         = [ 4.70-2.262 \times {\frac{1.83}{\sqrt{10} } } , 4.70+ 2.262 \times {\frac{1.83}{\sqrt{10} } } ]

                                         = [$3.39, $6.01]

Therefore, a 95% confidence interval for the true mean is [$3.39, $6.01].

The interpretation of the above result is that we are 95% confident that the true mean will lie between incomes of $3.39 and $6.01.

Also, the margin of error  =  2.262 \times {\frac{s}{\sqrt{n} } }

                                          =  2.262 \times {\frac{1.83}{\sqrt{10} } }  = <u>1.31</u>

4 0
3 years ago
7divided by 43.4 pls help
Vikki [24]

0.161      (they said explain it so)                                                                                                                                                                                                                                                                              

8 0
3 years ago
A motorboat travels 275 kilometers in 5 hours going upstream. It travels 405 kilometers going downstream in the same amount of t
VladimirAG [237]

Step-by-step explanation:

275 \times 5 = 1375

The rate of the motorboat on still water is 1375

405 \times 5 = 2025

The current rate of the motorboat is 2025

4 0
3 years ago
A mountain bike originally priced at $960.00 is up for sale at a discount of 14%. What is the price of the mountain bike after t
PilotLPTM [1.2K]

Answer:

$825.60

Step-by-step explanation:

Since the price of it is $960.00, it means that it's 100%.

100% = 960

1% = 960 ÷ 100 = 9.6

14% = 9.6 x 14 = 134.40

Since they asked for after the discount, deduct the price of the discount with the original price.

Mountain bike after 14% discount =

960 - 134.40 = 825.60

7 0
3 years ago
Read 2 more answers
Help please No links And with explanation please!
victus00 [196]
First: -8 second:17
4 0
3 years ago
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