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vichka [17]
3 years ago
11

A construction crew is lengthening a road. The road started with a length of 59 miles, and the crew is adding 2 miles to the roa

d each day.
Let L represent the total length of the road (in miles), and let D represent the number of days the crew has worked. Write an equation relating I to
this equation to find the total length of the road after the crew has worked 32 days.
Mathematics
1 answer:
Illusion [34]3 years ago
7 0
The answer is L = 59 + 2(d). If we plug in our numbers, we will get L = 59 + 2(32). 2*32 is 64, and if you add 59 you get 123 miles.

This isn’t apart of the question, but 59 would be the y-intercept. 2 miles per day (2/1) is the slope.
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Turn 15/6 into a mixed number​
laila [671]

Answer:

2 3/6

Step-by-step explanation:

divide 15 by 6 to get 2, and left with a remainder of 3.

6 0
3 years ago
A simple random sample of size nequals81 is obtained from a population with mu equals 83 and sigma equals 27. ​(a) Describe the
Ivanshal [37]

Answer:

a) \bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})

With:

\mu_{\bar X}= 83

\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3

b) z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2

P(Z>2) = 1-P(Z

c) z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45

P(Z

d) z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1

z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2

P(-1.2

Step-by-step explanation:

For this case we know the following propoertis for the random variable X

\mu = 83, \sigma = 27

We select a sample size of n = 81

Part a

Since the sample size is large enough we can use the central limit distribution and the distribution for the sampel mean on this case would be:

\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})

With:

\mu_{\bar X}= 83

\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3

Part b

We want this probability:

P(\bar X>89)

We can use the z score formula given by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And if we find the z score for 89 we got:

z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2

P(Z>2) = 1-P(Z

Part c

P(\bar X

We can use the z score formula given by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And if we find the z score for 75.65 we got:

z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45

P(Z

Part d

We want this probability:

P(79.4 < \bar X < 89.3)

We find the z scores:

z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1

z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2

P(-1.2

8 0
4 years ago
Is the GCF of 28 and 50
Georgia [21]
28: 1 , 2 , 4 , 7 , 14 , 28

50: 1 , 2 , 5 , 10 , 25 , 50

The GCF of 28 and 50 is 2.

----------------------------------------------------------

I hope that helps you!! Any more questions, please feel free to ask me!!
6 0
3 years ago
Read 2 more answers
On the first january 2014 carol invested some money in a bank account the account payes 2.5% compound interest per year on 1st j
Sladkaya [172]

Answer:

$23,360

Step-by-step explanation:

Calculation to determine how much carol originally invested in the account

First step is to divide £23517.60 by 1.025

= (23,517.60)/(1+.025)

= (23,517.60)/1.025

=$22,944

Second step is to add back the $1,000 withdrew

=$22,944+$1,000

=$23,944

Now let calculate how much carol originally invested in the account

$23,944=1.025P

Divide both side by 1.025

P=$23,944/1.025

P=$23,360

Therefore the amount that carol originally invested in the account is $23,360

4 0
3 years ago
Are the angles complementary, supplementary, or neither? What's the answer? Photo below.
soldier1979 [14.2K]
I personally would say neither because they either aren't the same or there not equivalent.
3 0
3 years ago
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