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3 years ago

Help plz I don’t get it

Mathematics

2 answers:

AleksandrR [38]3 years ago

8 0

1-a-1 13
1-b-1 2 3 5 6 10 15 30

laila [671]3 years ago

3 0

(I will do the first two parts of each question)

1a. 13 is only divisible by 1 and 13, so they are its factors.

1b. 30 can be divided by 1 and 30, and 2*15=30, 3*10=30, and 5*6=30, so 30's factors are 1, 2, 3, 5, 6, 10, 15, and 30.

2. 75's factors are: 1, 3, 5, 15, 25, and 75. So, there can be 75 piles of 1, 25 piles of 3, 15 piles of 5, etc. (3 more)

3a. 3*1=3, 3*2=6, 3*3=9, etc. so the first 5 multiples of 3 are: 3, 6, 9, 12, 15

3b. Same as before, so: 10, 20, 30, 40, 50

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What is the value of the rationale expression below when x is equal to 4? X+20/x+4

sveta [45]

Answer:

Step-by-step explanation:

to evaluate substitute x = 4 into the rational expression, that is

$\frac{4+20}{4+4}$ = $\frac{24}{8}$ = 3

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3 years ago

Suppose a geyser has a mean time between irruption’s of 75 minutes. If the interval of time between the eruption is normally dis

lesya [120]

Answer:

(a) The probability that a randomly selected Time interval between irruption is longer than 84 minutes is 0.3264.

(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.

(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.

(d) The probability decreases because the variability in the sample mean decreases as we increase the sample size

(e) The population mean may be larger than 75 minutes between irruption.

Step-by-step explanation:

We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.

(a) Let X = the interval of time between the eruption

So, X ~ Normal( $\mu=75, \sigma^{2} =20$ )

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 between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

Now, the probability that a randomly selected Time interval between irruption is longer than 84 minutes is given by = P(X > 84 min)

P(X > 84 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{84-75}{20}$ ) = P(Z > 0.45) = 1 - P(Z $\leq$ 0.45)

= 1 - 0.6736 = 0.3264

The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.

(b) Let $\bar X$ = sample time intervals between the eruption

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 between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P( $\bar X$ > 84 min)

P( $\bar X$ > 84 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{84-75}{\frac{20}{\sqrt{13} } }$ ) = P(Z > 1.62) = 1 - P(Z $\leq$ 1.62)

= 1 - 0.9474 = 0.0526

The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.

(c) Let $\bar X$ = sample time intervals between the eruption

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 between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

n = sample of time intervals = 20

Now, the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is given by = P( $\bar X$ > 84 min)

P( $\bar X$ > 84 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{84-75}{\frac{20}{\sqrt{20} } }$ ) = P(Z > 2.01) = 1 - P(Z $\leq$ 2.01)

= 1 - 0.9778 = 0.0222

The above probability is calculated by looking at the value of x = 2.01 in the z table which has an area of 0.9778.

(d) When increasing the sample size, the probability decreases because the variability in the sample mean decreases as we increase the sample size which we can clearly see in part (b) and (c) of the question.

(e) Since it is clear that the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is very slow(less than 5%0 which means that this is an unusual event. So, we can conclude that the population mean may be larger than 75 minutes between irruption.

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3 years ago

why do whole numbers raised to an exponet get greater while fractions raised to an exponet get smaller

hoa [83]

Fractions are basically divisions.
Ex. 2/4 = 1/2

And whole numbers like 10, get greater when raised to an exponent.
Ex. 10^2 = 100

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3 years ago

Please complete these with explanation

Leya [2.2K]

Answer:

minus 5

Step-by-step explanation:

I think so it might be not a correct answer

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3 years ago

Read 2 more answers

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zvonat [6]

Answer:

See below.

Step-by-step explanation:

This is how you prove it.

<B and <F are given as congruent.

This is 1 pair of congruent angles for triangles ABC and GFE.

<DEC and <DCE are given as congruent.

Using vertical angles and substitution of transitivity of congruence of angles, show that angles ACB and GEF are congruent.

This is 1 pair of congruent angles for triangles ABC and GFE.

Now you need another side to do either AAS or ASA.

Look at triangle DCE. Using the fact that angles DEC and DCE are congruent, opposite sides are congruent, so segments DC and DE are congruent. You are told segments DF and BD are congruent. Using segment addition postulate and substitution, show that segments CB and EF are congruent.

Now you have 1 pair of included sides congruent ABC and GFE.

Now using ASA, you prove triangles ABC and GFE congruent.

7 0

3 years ago