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Shalnov [3]
3 years ago
9

HELP PLEASE IM GONNA FAIL21. Identify as a function or not a Function ​

Mathematics
1 answer:
Vladimir [108]3 years ago
5 0

Answer:

First one: function

Second one: not a function

Step-by-step explanation:

A function is a relationship where each input has it's own unique output. If an input has more than one output then the relationship is not a function

For the first one no x value repeats meaning each input has it's own output therefore it is a function

For the second one the x value 8 has more than 1 output therefore the second one is not a function

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what is the completely factored form of xy3 – x3y? (the 3's are powers if it wasn't clear.)a. xy(y x)(y – x)b. xy(y – x)(y – x)c
VARVARA [1.3K]
I think the correct answer from the choices listed above is option A. The completely factored formed of the function xy^3 – (x^3)y would be xy (x+y) (x-y). It is factored as follows:

<span>xy^3 – x^3y
xy (y^2 - x^2)
xy (x+y) (x-y)
</span><span>
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7 0
3 years ago
Read 2 more answers
Cos X<br> Y<br> 12<br> X<br> 16<br> 20<br> Z<br> 4<br> A) 3/4<br> B) 4/3<br> C)3/5<br> D) 5/4
yanalaym [24]

Answer:

C.

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5 0
3 years ago
10∙20=8∙d210•20=8•d2 .
tekilochka [14]
Is that the whole equation? Exactly from the book or paper?

5 0
4 years ago
Suppose a geyser has a mean time between eruptions of 72 minutes. Let the interval of time between the eruptions be normally dis
nikitadnepr [17]

Answer:

(a) The probability that a randomly selected time interval between eruptions is longer than 82 ​minutes is 0.3336.

(b) The probability that a random sample of 13-time intervals between eruptions has a mean longer than 82 ​minutes is 0.0582.

(c) The probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 ​minutes is 0.0055.

(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.

(e) The population mean must be more than 72​, since the probability is so low.

Step-by-step explanation:

We are given that a geyser has a mean time between eruptions of 72 minutes.

Also, the interval of time between the eruptions be normally distributed with a standard deviation of 23 minutes.

(a) Let X = <u><em>the interval of time between the eruptions</em></u>

So, X ~ N(\mu=72, \sigma^{2} =23^{2})

The z-score probability distribution for the normal distribution is given by;

                            Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean time = 72 minutes

           \sigma = standard deviation = 23 minutes

Now, the probability that a randomly selected time interval between eruptions is longer than 82 ​minutes is given by = P(X > 82 min)

       P(X > 82 min) = P( \frac{X-\mu}{\sigma} > \frac{82-72}{23} ) = P(Z > 0.43) = 1 - P(Z \leq 0.43)

                                                           = 1 - 0.6664 = <u>0.3336</u>

The above probability is calculated by looking at the value of x = 0.43 in the z table which has an area of 0.6664.

(b) Let \bar X = <u><em>sample mean time between the eruptions</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time = 72 minutes

           \sigma = standard deviation = 23 minutes

           n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between eruptions has a mean longer than 82 ​minutes is given by = P(\bar X > 82 min)

       P(\bar X > 82 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{82-72}{\frac{23}{\sqrt{13} } } ) = P(Z > 1.57) = 1 - P(Z \leq 1.57)

                                                           = 1 - 0.9418 = <u>0.0582</u>

The above probability is calculated by looking at the value of x = 1.57 in the z table which has an area of 0.9418.

(c) Let \bar X = <u><em>sample mean time between the eruptions</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time = 72 minutes

           \sigma = standard deviation = 23 minutes

           n = sample of time intervals = 34

Now, the probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 ​minutes is given by = P(\bar X > 82 min)

       P(\bar X > 82 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{82-72}{\frac{23}{\sqrt{34} } } ) = P(Z > 2.54) = 1 - P(Z \leq 2.54)

                                                           = 1 - 0.9945 = <u>0.0055</u>

The above probability is calculated by looking at the value of x = 2.54 in the z table which has an area of 0.9945.

(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.

(e) If a random sample of 34-time intervals between eruptions has a mean longer than 82 ​minutes, then we conclude that the population mean must be more than 72​, since the probability is so low.

6 0
3 years ago
Please help me and ill mark you as brainlest ;)!
JulsSmile [24]
The hypotenuse would be 9 inches as 6 is equal to 2/3 of 9 :)
6 0
3 years ago
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