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2 years ago

PLEASE HELP WILL GIVE BRAINLIEST

Mathematics

1 answer:

lora16 [44]2 years ago

5 0

Answer:

B. y = 3x + 1

Step-by-step explanation:

We can look at the table like a bunch of values, where (x) is what is inserted into the equation, and (y) is what comes out.

Now, simply insert the (x) values into each equation and see if they work out or not.

y = (1/3)(-1) + 1 does not equal -2, meaning this equation is defunct.

y = 3(-1) + 1 = -2; y = 3(0) + 1 = 1. This must be the equation as it works for the (x) values given. You can try the other equations, but a simple look at them is all you need to prove them false.

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Question #2 What is the value of the expression 35 ? 23 – 1 A 5 B 7 C 12

algol13

Answer:

A. 5

Step-by-step explanation:

just use calculator tbh

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3 years ago

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XD me banearon de nuevo pero regrese igualmente no me pueden moderadores.

aleksandrvk [35]

Answer:

yeah what you said

7 0

3 years ago

Help me with this please

drek231 [11]

The answer is only C and D

6 0

3 years ago

Assume that females have pulse rates that are normally distributed with a mean of μ=73.0 beats per minute and a standard deviati

Gennadij [26K]

Answer:

a. the probability that her pulse rate is less than 76 beats per minute is 0.5948

b. If 25 adult females are randomly selected, the probability that they have pulse rates with a mean less than 76 beats per minute is 0.8849

c. D. Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

Step-by-step explanation:

Given that:

Mean μ =73.0

Standard deviation σ =12.5

a. If 1 adult female is randomly selected, find the probability that her pulse rate is less than 76 beats per minute.

Let X represent the random variable that is normally distributed with a mean of 73.0 beats per minute and a standard deviation of 12.5 beats per minute.

Then : X $\sim$ N ( μ = 73.0 , σ = 12.5)

The probability that her pulse rate is less than 76 beats per minute can be computed as:

$P(X < 76) = P(\dfrac{X-\mu}{\sigma}< \dfrac{X-\mu}{\sigma})$

$P(X < 76) = P(\dfrac{76-\mu}{\sigma}< \dfrac{76-73}{12.5})$

$P(X < 76) = P(Z< \dfrac{3}{12.5})$

$P(X < 76) = P(Z< 0.24)$

From the standard normal distribution tables,

$P(X < 76) = 0.5948$

Therefore , the probability that her pulse rate is less than 76 beats per minute is 0.5948

b. If 25 adult females are randomly selected, find the probability that they have pulse rates with a mean less than 76 beats per minute.

now; we have a sample size n = 25

The probability can now be calculated as follows:

$P(\overline X < 76) = P(\dfrac{\overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{ \overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}})$

$P( \overline X < 76) = P(\dfrac{76-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{76-73}{\dfrac{12.5}{\sqrt{25}}})$

$P( \overline X < 76) = P(Z< \dfrac{3}{\dfrac{12.5}{5}})$

$P( \overline X < 76) = P(Z< 1.2)$

From the standard normal distribution tables,

$P(\overline X < 76) = 0.8849$

c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

In order to determine the probability in part (b); the normal distribution is perfect to be used here even when the sample size does not exceed 30.

Therefore option D is correct.

Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

5 0

2 years ago

A neon light flashes five times every 10 seconds. Show that the light flashes 43 200 times in one day.

11Alexandr11 [23.1K]

1 day = 24 hours

1 hour = 60 minutes

1 minute = 60 seconds

60 seconds x 60 minutes = 3,600 seconds per hour

3,600 seconds per hour x 24 hours = 86,400 seconds per day.

The light flashes 5 times every 10 seconds:

5 flashes / 10 seconds = 1 flash every 2 seconds

86,400 seconds / 2 seconds = 43,200 flashes per day.

4 0

2 years ago