Someone please explain the answer to me

Mathematics

1 answer:

tia_tia [17]4 years ago

7 0

You can divide the fractions

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I need help asap, will give brainliest and 50 points!!

hodyreva [135]

When f(x) = g(x), it is the coordinates when the two graphs intersected.

The answer is read off from the graph.

Answer: x = -3.5 and x = -0.5 (Answer 1)

6 0

4 years ago

A board 91 cm long is to be cut into two parts so that the longer part is 10 cm more than twice as long as the shorter part. Fin

mojhsa [17]

Answer:

The parts are 27 cm and 64 cm long.

Step-by-step explanation:

Let the shorter part be x.

"The longer part is 10 cm more than twice as long as the shorter part."

shorter part: x

longer part: 2x + 10

total length: x + 2x + 10

total length: 91

x + 2x + 10 = 91

3x + 10 = 91

3x = 81

x = 27

2x + 10 = 2(27) + 10 = 54 + 10 = 64

Answer: The parts are 27 cm and 64 cm long.

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3 years ago

Add. 4 7/12 + 2 3/4 + 7 5/12 Help

natta225 [31]

Answer:

14 3/4 hope this helps

Brainliest?

Step-by-step explanation:

6 0

4 years ago

Read 2 more answers

Part of a university's professor's job is to publish his or her research. This task often entails reading a variety of journal a

goblinko [34]

Answer:

(18.1042, 35.2292) is a 90% confidence interval for the mean number of journal articles read monthly by professors.

Step-by-step explanation:

We have a small sample of size n = 12, $\bar{x} = 26.6667$ and s = 16.5163. A pivotal quantity for this case is given by $T = \frac{\bar{X}-\mu}{S/\sqrt{n}}$ which has a t distribution with n-1 degrees of freedom if we suppose that we take the sample from the normal population. The confidence interval is given by $\bar{x}\pm t_{\alpha/2}(\frac{s}{\sqrt{n}})$ where $t_{\alpha/2}$ is the $\alpha/2$ th quantile of the t distribution with n - 1 = 12 - 1 = 11 degrees of freedom. As we want the 90% confidence interval, we have that $\alpha = 0.1$ and the confidence interval is $26.6667\pm t_{0.05}(\frac{16.5163}{\sqrt{12}})$ where $t_{0.05}$ is the 5th quantile of the t distribution with 11 df, i.e., $t_{0.05} = -1.7959$ . Then, we have $26.6667\pm (-1.7959)(\frac{16.5163}{\sqrt{12}})$ and the 90% confidence interval is given by (18.1042, 35.2292).