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Brian has DVDs and CDs. The number of CDs he has can be modelled with the formula C = 2D + 11, where C represents the number of

wel

Answer: 15 DVDs

Step-by-step explanation:

C = 2D + 11

41 = 2D + 11

2D = 30

D = 15 DVDs

7 0

2 years ago

Hilda was simplifying some numerical expressions and made each of the following sequences of calculations. Name the mathematical

Galina-37 [17]

Answer:

Problem 1:

5 · (−4/3) · (2/5)

= (−4/3) · 5 · (2/5) Conmutative Property of Multiplication

= (−4/3) · (5 · (2/5)) Associative Property of Multiplication

= (−4/3) · (2/1) = −8/3 = −2*2/3 Multiplied fractions and extracted common factor

Problem 2:

17 + 29 + 3+ 1

= 17 + 3 + 29 + 1 Conmutative Property of Addition

= (17 + 3) + (29 + 1) Associative Property of Addition

= 20 + 30 Added groups

= 50 Added terms

Step-by-step explanation:

For the Problem 1:

In the first step, Hilda applied the Conmutative Property of Multiplication, because she changed the order of the numbers in the product

In the second step, she applied the Associative Property of Multiplication, because she agrouped the product of 5 and 2/5 to perform it sepparately

In the third step, she calculated the product of the fractions -4/3 and 2/1, then she extracted 2 as a common factor to express the fraction as -2*2/3

For the Problem 2:

In the first step, Hilda applied the Conmutative Property of Addition, because she changed the order of the numbers in the sum

In the second step, she applied the Associative Property of Addition, because she associated the addition of 17 and 3 and the addition of 29 and 1, to calculate them in groups.

4 0

3 years ago

Read 2 more answers

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.

8 0

3 years ago

If h(x) = 6 – x, what is the value of (hxh) (10)

vova2212 [387]

For those who may need it in the future, the answer on e1010 is actually: C: 10

8 0

3 years ago

36-3•4 15-9 Divided by three

matrenka [14]

Answer:

10

Step-by-step explanation: