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xz_007 [3.2K]
3 years ago
10

How much is 4817 times 818?

Mathematics
2 answers:
Free_Kalibri [48]3 years ago
8 0
3940306 im new in this app, hey guys !
Nadusha1986 [10]3 years ago
7 0
3940306

set it up like this 818*4817 = 818*(4000+800+10+7) = 818*4000+818*800+818*10+818*7 
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a craft store retailer is using a discount rate of %40 on Christmas scrapbook items if a package of scrapbook paper cost 15.59 w
tester [92]
38.98 because you get 15.59 and divide that by 40% you get 38.975 but since you're speaking in money it rounds to 38.98
6 0
3 years ago
Need help asap this is due in 20 min!!! i will gove brainliest too! questions 8-11
andreev551 [17]

8. x= 7

9. x= 9

10. x= -10

11. x= -18

7 0
3 years ago
36 out of 100 randomly selected taxpayers knew about tax incentives for installing energy-saving furnaces. Find a 90% confidence
tresset_1 [31]

Answer:

The 90% confidence interval for the population proportion who knew about the incentives is (0.28, 0.44).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

n = 100, \pi = \frac{36}{60} = 0.6

90% confidence level

So \alpha = 0.1, z is the value of Z that has a pvalue of 1 - \frac{0.1}{2} = 0.95, so Z = 1.645.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.36 - 1.645\sqrt{\frac{0.36*0.64}{100}} = 0.28

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.36 + 1.645\sqrt{\frac{0.36*0.64}{100}} = 0.44

The 90% confidence interval for the population proportion who knew about the incentives is (0.28, 0.44).

7 0
3 years ago
solve for each please i really need help if u want to help me with my test and i get an a or a b i will give u 500 dollars
Len [333]

Answer:

The relation is not a function

The domain is {1, 2, 3}

The range is {3, 4, 5}

Step-by-step explanation:

A relation of a set of ordered pairs x and y is a function if

  • Every x has only one value of y
  • x appears once in ordered pairs

<u><em>Examples:</em></u>

  • The relation {(1, 2), (-2, 3), (4, 5)} is a function because every x has only one value of y (x = 1 has y = 2, x = -2 has y = 3, x = 4 has y = 5)
  • The relation {(1, 2), (-2, 3), (1, 5)} is not a function because one x has two values of y (x = 1 has values of y = 2 and 5)
  • The domain is the set of values of x
  • The range is the set of values of y

Let us solve the question

∵ The relation = {(1, 3), (2, 3), (3, 4), (2, 5)}

∵ x = 1 has y = 3

∵ x = 2 has y = 3

∵ x = 3 has y = 4

∵ x = 2 has y = 5

→ One x appears twice in the ordered pairs

∵ x = 2 has y = 3 and 5

∴ The relation is not a function because one x has two values of y

∵ The domain is the set of values of x

∴ The domain = {1, 2, 3}

∵ The range is the set of values of y

∴ The range = {3, 4, 5}

3 0
3 years ago
The following equation will have two solutions, y = x2 - 2x - 15?<br> True<br> False
motikmotik
The answer would be False.
3 0
3 years ago
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