I need help ASAP Thankyou!

What is the range of the function shown below?

Ivenika [448]

Answer:

Step-by-step explanation:

{ - 9, - 4, - 1}

5 0

3 years ago

In 2012 your car was worth $10,000. In 2014 your car was worth $8,800. Suppose the value of your car decreased at a constant rat

Sonbull [250]

Answer:

g(t) = 10000(0.938)^t

Step-by-step explanation:

Given data:

car worth is $10,000 in 2012

car worth is $8000 in 2014

let linear function is given as

P(t) = at + b

which denote the value of car in year t

take t =0 for year 2012

at t =0, 10,000 = 0 + b

we get b = 10,000

take t =2 for year 2014

at t =2, P(2) = 2a + b

8800 = 2a + 10,000

a = - 600

Thus the price of car at year t after 2012 is given as p(t) = -600t + 10000

let the exponential function $p(t) =ab^2$ where t denote t = 0 at 2012

putting t = 0 P(0) = 10,000 we get 10,000 = ab^0

a = 10,000

putting t = 2 p = 8800

$8800 =ab^2$

$b^2 = \frac{8800}{10000}$

b = 0.938

g(t) = 10000(0.938)^t

3 0

3 years ago

Determine whether the value is a discrete random variable, continuous random variable, or not a random variable.

gulaghasi [49]

As a rule of thumb, a discrete random variable is countable.
A continuous random variable is measurable but not countable.

a. Discrete
The number of hits to a website is countable.

b. Continuous
The weight of a T-bone steak is measurable.

c. Discrete
The political party affiliations of adults are countable.

d. Discrete
The number of bald eagles in a country is countable.

e. Continuous
The amount of snowfall in December is measurable.

f. Discrete
The number of textbook authors is countable.

6 0

3 years ago

Read 2 more answers

A report indicated that 37% of adults had received a bogus email intended to steal personal information. Suppose a random sample

fiasKO [112]

Answer:

5.05% probability that no more than 34% had received such an email.

Step-by-step explanation:

We use the binomial approximation to the normal to solve this problem.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .