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4 years ago

I'm currently in the 4th grade and working on one of my assignments. I'm not too sure how to solve this problem:

2 answers:

7 0

So all i did was start adding them together.

you will have 3 cds that cost $11.95 and 6 cds that cost $12.75

$11.95 + 11.95+11.95 + 12.75+12.75+12.75+12.75+12.75+12.75= 112.35$

saveliy_v [14]4 years ago

6 0

ad all the Cds and u will get ur answer so add ur numbers with 9

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Simplify: 2⁴ x 10³ (-5)³ x (-2)

Crazy boy [7]

Answer:

hope this helps

Step-by-step explanation:

7 0

3 years ago

Find the value. 0.003 0.03 0.3 3

pickupchik [31]

｡☆✼★ ━━━━━━━━━━━━━━━ ☾

The correct option would be C. 0.3

You can test this:

0.3^3 = 0.027

Have A Nice Day ❤

Stay Brainly! ヅ

- Ally ✧

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7 0

3 years ago

Police response time to an emergency call is the difference between the time the call is first received by the dispatcher and th

andrew11 [14]

Answer:

a) 0.7683 = 76.83% probability that a randomly selected emergency call is between 5 and 10 minutes.

b) 0.0606 = 6.06% probability that a randomly received emergency call is of less than 5 min.

c) 0.1711 = 17.11% probability that a randomly received emergency call is of more than 10 min.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 8.1 minutes and a standard deviation of 2.0 minutes.

This means that $\mu = 8.1, \sigma = 2$

a. between 5 and 10 min

This is the p-value of Z when X = 10 subtracted by the p-value of Z when X = 5.

X = 10

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{10 - 8.1}{2}$

$Z = 0.95$

$Z = 0.95$ has a p-value of 0.8289

X = 5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5 - 8.1}{2}$

$Z = -1.55$

$Z = -1.55$ has a p-value of 0.0606

0.8289 - 0.0606 = 0.7683

0.7683 = 76.83% probability that a randomly selected emergency call is between 5 and 10 minutes.

b. less than 5 min

p-value of Z when X = 5, which, found from item a, is of 0.0606

0.0606 = 6.06% probability that a randomly received emergency call is of less than 5 min.

c. more than 10 min

1 subtracted by the p-value of Z when X = 10, which, from item a, is of 0.8289

1 - 0.8289 = 0.1711

0.1711 = 17.11% probability that a randomly received emergency call is of more than 10 min.

7 0

3 years ago

The time between telephone calls to a cable television service call center follows an exponential distribution with a mean of 1.

Ulleksa [173]

Answer:

0.52763 is the probability that the time between the next two calls will be 54 seconds or less.

0.19285 is the probability that the time between the next two calls will be greater than 118.5 seconds.

Step-by-step explanation:

We are given the following information in the question:

The time between telephone calls to a cable television service call center follows an exponential distribution with a mean of 1.2 minutes.

The distribution function can be written as:

$f(x) = \lambda e^{-\lambda x}\\\text{where lambda is the parameter}\\\\\text{Mean} = \mu = \displaystyle\frac{1}{\lambda}\\\\\Rightarrow 1.2 = \frac{1}{\lambda}\\\\\lambda = 0.84 \\f(x) = 0.84 e^{0.84 x}$

The probability for exponential distribution is given as:

$P( x \leq a) = 1 - e^{\frac{-a}{\mu}}\\\\P(a \leq x \leq b) = e^{\frac{-a}{\mu} -\frac{-b}{\mu}}$

a) P( time between the next two calls will be 54 seconds or less)

$P( x \leq 0.9)\\= 1 - e^{\frac{\frac{-54}{60}}{1.2}} = 0.52763$

0.52763 is the probability that the time between the next two calls will be 54 seconds or less.

b) P(time between the next two calls will be greater than 118.5 seconds)

$p( x > \frac{118.5}{60}) = P(x > 1.975)\\\\ = 1 - P(x \leq 1.975) \\\\= 1 -1+ e^{\frac{-1.975}{1.2}}\\\\= 0.19285$

0.19285 is the probability that the time between the next two calls will be greater than 118.5 seconds.

6 0

4 years ago

Helppppppppppppppppp

gavmur [86]

I believe a is the second and fourth options and b is the first and third

6 0

3 years ago