(x^3 + 5)(2x – 1)<br>(x^2

What function is the inverse of the exponential function y = (1.5)x?

vagabundo [1.1K]

Correct question is;

What function is the inverse of the exponential function y = 1.5^(x)?

Answer:

y = log_1.5_x

Step-by-step explanation:

The inverse of exponential functions is usually written in form of logarithm.

For example inverse of y = p^(x) will be written as; y = log_p_(x)

Similarly applying this same pattern to our exponential function y = 1.5^(x), we have the inverse as;

y = log_1.5_x

6 0

2 years ago

A bowling leagues mean score is 197 with a standard deviation of 12. The scores are normally distributed. What is the probabilit

Bond [772]

Answer:

0.281 = 28.1% probability a given player averaged less than 190.

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.

A bowling leagues mean score is 197 with a standard deviation of 12.

This means that $\mu = 197, \sigma = 12$

What is the probability a given player averaged less than 190?

This is the p-value of Z when X = 190.

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

$Z = \frac{190 - 197}{12}$

$Z = -0.58$

$Z = -0.58$ has a p-value of 0.281.

0.281 = 28.1% probability a given player averaged less than 190.

8 0

3 years ago

The population of a certain animal species you are studying decreases at a rate of 3.5% per year. Only 80 of the animals in the

balu736 [363]

The function:
f ( x ) = x * 0.965^t
where x is the initial amount and t is the number of the years
80 = x * 0.965^3
80 = x * 0.89632
x = 80 : 0.89632 ≈ 89
Answer:
The initial amount of animals was 89.

3 0

3 years ago

0.4y − 0.2y + 10 = 11.8

mixer [17]

Answer:

0.2y+10

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

A Normal model states that if we draw repeated random samples of the same size, n, from some population and measure sample pro

notka56 [123]

Answer:

1) Randomization condition: We assume that we are selecting random samples so this condition is satisfied

2) 10% condition: We assume that the random sample selected is less than 10% of the population size

3) Success/ Failure condition:

For this case we need to satisfy this:

$np \geq 10, n(1-p)\geq 10$

So since the condition 1 and 2 are satisfied the correct option for this case would be:

a)np and nq must be respectively at least equal to 10

Step-by-step explanation:

Assuming the following options:

a)np and nq must be respectively at least equal to 10

b) as n increases, the distribution of sample proportions becomes less Normal

c) np and nq must be respectively less than 10

d) np plus nq must be at least 10.

For this case we need 3 basic conditions:

1) Randomization condition: We assume that we are selecting random samples so this condition is satisfied

2) 10% condition: We assume that the random sample selected is less than 10% of the population size

3) Success/ Failure condition:

For this case we need to satisfy this:

$np \geq 10, n(1-p)\geq 10$

So since the condition 1 and 2 are satisfied the correct option for this case would be:

a)np and nq must be respectively at least equal to 10

8 0

3 years ago

(x^3 + 5)(2x – 1)(x^2 – 4)(x^2 + 4)​

(x^3 + 5)(2x – 1)(x^2 – 4)(x^2 + 4)