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3 years ago

What is the domain of this function? (8,8) (2, 19) (0,9) (-12, 19)

Mathematics

1 answer:

melamori03 [73]3 years ago

5 0

Answer:

The domain is -12, 0, 2, 8.

Step-by-step explanation:

The domain is the set of all possible x-values. The x-values are the first numbers in a point.

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Need help please thank you if you help

Alexus [3.1K]

Answer:

$1) 3\f\frac{2}{3}=2\\2) x\frac{x}{18} =18\\3)12\frac{12}{y}=y \\4)x\frac{x}{6} =6$

7 0

2 years ago

Can you answer this one

qaws [65]

Answer:

P'(-6,-4)

Step-by-step explanation:

By translating the point under the translation, to put it into a simpler form, it says to move the point 4 units to the left and 6 units down. Using this, the point will move to P'(-6,-4)

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3 years ago

Read 2 more answers

Vivi is a drummer for a band. She burns 756756756 calories while drumming for 333 hours. She burns the same number of calories e

alexandr1967 [171]

burned: 756 calories

hours: 3

same number of calories each hour

-> to find this, we must split the number of calories evenly among the hours

steps

756/3

= 252 calories per hour

6 0

3 years ago

Read 2 more answers

A card is chosen at random from a standard deck.

allsm [11]

Answer:

6/13

Step-by-step explanation:

There are 52 cards in a deck, and 26 of them are red. There are 2 red queens in a deck, so the probability of choosing a red card that is not a queen is 24/52 = 6/13.

7 0

2 years ago

Read 2 more answers

The contents of a sample of 26 cans of apple juice showed a standard deviation of 0.06 ounces. We are interested in testing to d

NikAS [45]

Answer:

Option b. should not be rejected

Step-by-step explanation:

We are given that the contents of a sample of 26 cans of apple juice showed a standard deviation of 0.06 ounces.

We have to test whether the variance of the population is significantly more than 0.003, i.e.;

Null Hypothesis, $H_0$ : $\sigma$ = $\sqrt{0.003}$

Alternate Hypothesis, $H_1$ : $\sigma > \sqrt{0.003}$

The test statistics used here for testing variance is;

T.S. = $\frac{(n-1)s^{2} }{\sigma^{2} }$ ~ $\chi^{2}__n_-_1$

where, s = sample standard deviation = 0.06

n = sample size = 26 cans

So, Test statistics = $\frac{(26-1)0.06^{2} }{0.003 }$ ~ $\chi^{2}__2_5$

= 30

So, at 5% level of significance chi square table gives critical value of 37.65 at 25 degree of freedom. Since our test statistics is less than the critical so we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that null hypothesis should not be rejected and variance of population is 0.003.

5 0

3 years ago