All categories

2 years ago

For the function above, what is the missing value in the table below:

Mathematics

1 answer:

ollegr [7]2 years ago

7 0

Answer- 4
U need to plug in 1 for x and solve which would be 4

You might be interested in

The perimeter of a rectangle is 400 feet and the length of the rectangle is 3 times the width. find the dimensions of the rectan

bazaltina [42]

Width: x
Length: 3x
Perimeter=x+x+3x+3x=8x
8x=400
x=50
Dimension is 150×50 feet (length × width)

8 0

3 years ago

I need help on this please

tia_tia [17]

9514 1404 393

Answer:

B 2/6

Step-by-step explanation:

2 of the 6 possible outcomes are ones that are of interest. A "fair" die means the mutually-exclusive outcomes have equal probability, so ...

P(4 or 5) = P(4) +P(5) = 1/6 + 1/6 = 2/6

3 0

2 years ago

NASA is conducting an experiment to find out the fraction of people who black out at G forces greater than 6. In an earlier stud

SSSSS [86.1K]

Answer:

The large sample n = 190.44≅190

The large sample would be required in order to estimate the fraction of people who black out at 6 or more Gs at the 85% confidence level with an error of at most 0.04 is n = 190.44

<u>Step-by-step explanation</u>:

Given population proportion was estimated to be 0.3

p = 0.3

Given maximum of error E = 0.04

we know that maximum error

$M.E = \frac{Z_{\alpha } \sqrt{p(1-p)} }{\sqrt{n} }$

The 85% confidence level $z_{\alpha } = 1.44$

$\sqrt{n} = \frac{Z_{\alpha } \sqrt{p(1-p)} }{m.E}$

$\sqrt{n} = \frac{1.44X\sqrt{0.3(1-0.3} }{0.04}$

now calculation , we get

√n=13.80

now squaring on both sides n = 190.44

large sample n = 190.44≅190

<u>Conclusion</u>:-

Hence The large sample would be required in order to estimate the fraction of people who black out at 6 or more Gs at the 85% confidence level with an error of at most 0.04 is n = 190.44

7 0

2 years ago

20.) Using the cross multiplication method, which of the following is the solution to 18/6x = 4/(x+3)?

Likurg_2 [28]

Answer:

B. 9

Step-by-step explanation:

7 0

2 years ago

In a coin flipping game, a prize is won if heads is thrown 45%-55% of the time. In a second game, a prize is won if heads is thr

erica [24]

You would rather flip the coin 500 times in the first game because over time the percentage should average out to that range.

You would flip the coin 25 times in the second game to let outliers have a greater chance of making a dramatic difference, like having the average be strayed from 50:50 to 25:75 or less

4 0

2 years ago

Read 2 more answers