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labwork [276]
3 years ago
13

How do I solve this?

Mathematics
1 answer:
boyakko [2]3 years ago
3 0
See below for solution. Hope this is helpful

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Given: || EIJ GJI Prove: JIK IJL
EastWind [94]
Jk+jik+ijl goodluck :)

6 0
3 years ago
Please i need workings for these questions. Any kind of help will be appreciated.
Anna11 [10]

Answer:

1. N4 = a

2. N8.25 = b

Step-by-step explanation:

1000 litre =N3

Each additional 100 litres = 5 k

100 litres

2000 litres

100 litres =5k

2000 litres = x

100x =5×2000

100x =10,000

Divide both sides of the equation by 100

x = 100k

100 Kobo = N1

x =N1

N3+N1 =N4

2. First ten words cost N6

Each additional word cost 45 k

1= 45k

5=x

x = 5×45

= 225 Kobo

100 Kobo = 1 Naira

225 Kobo = x

x = N2.25

N6 + N2.25

x = N8.25 = B

4 0
3 years ago
Which type of function best models the data shown on the scatterplot?
Flauer [41]
Due to the u-shaped graph suggested by these data points, this would best be matched with a quadratic function.
5 0
4 years ago
Read 2 more answers
Which expression below gives the average rate of change of the function g(x) = -x2 - 4x on the interval 6 ≤ x ≤ 8 ?
Paha777 [63]

Answer: -18

<u>Explanation:</u>

Average rate of change is slope: \frac{y_{2}-y_{1}}{x_{2}-x_{1}}.  Need to find the y-coordinates:

g(x) = -x² - 4x

g(6) = -(6)² - 4(6)

      = -36 - 24

      = -60

⇒ (6, -60)

g(8) = -(8)² - 4(8)

      = -64 - 32

      = -96

⇒ (8, -96)

m = \frac{-96 - (-60)}{8 - 6}

   = \frac{-36}{2}

   = -18

7 0
3 years ago
Read 2 more answers
Find a 75% confidence interval for the population average weight μ of all adult mountain lions in the specified region. (Round y
kupik [55]

Answer:

75% confidence interval is 91.8±16.66. That is between 75.1 and 108.5 pounds.

Step-by-step explanation:

The question is missing. It is as follows:

Adult wild mountain lions (18 months or older) captured and released for the first time in the San Andres Mountains had the following weights (pounds): 69  104  125  129  60  64

Assume that the population of x values has an approximately normal distribution.

Find a 75% confidence interval for the population average weight μ of all adult mountain lions in the specified region. (Round your answers to one decimal place.)

75% Confidence Interval can be calculated using M±ME where

  • M is the sample mean weight of the wild mountain lions (\frac{69 +104 +125 +129+60 +64}{6} =91.8)
  • ME is the margin of error of the mean

And margin of error (ME) of the mean can be calculated using the formula

ME=\frac{t*s}{\sqrt{N} } where

  • t is the corresponding statistic in the 75% confidence level and 5 degrees of freedom (1.30)
  • s is the standard deviation of the sample(31.4)
  • N is the sample size (6)

Thus, ME=\frac{1.30*31.4}{\sqrt{6} } ≈16.66

Then 75% confidence interval is 91.8±16.66. That is between 75.1 and 108.5

3 0
3 years ago
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