I’m pretty sure you would have a 72% which would be a low C but maybe still passing. My schools passing grade is 70%. You would have to see what a passing grade is in your school.

4 0

2 years ago

Find f(-2) for f(x) = 2(4)^x

stepladder [879]

Answer:

2(4)^-2

=2/4^2

2/16

1/8

thanks

3 0

3 years ago

Read 2 more answers

A population of insects, in thousands, can be molded using the function (t)= 1.75(0.97)x

Dominik [7]

Question:

A population of insects, in thousands, can be modeled using the function

$p(t) = 1.75(0.97)^t$ , where t is time in months. Which statement best

describes the population of insects?

A. The population is decaying at a rate of 3% each month.

B. The population is decaying at a rate of 25% each month.

C. The population is growing at a rate of 75% each month.

D. The population is growing at a rate of 97% each month.

Answer:

A. The population is decaying at a rate of 3% each month.

Step-by-step explanation:

Given

$p(t) = 1.75(0.97)^t$

Required

True statement about the function

From the options, we can see that we are to answer the question on the basis of decay and growing rates.

An exponential form is:

$y=ab^x$

Compare to $p(t) = 1.75(0.97)^t$

$b= 0.97\\$

If $b > 1$ , then $b = 1 + r$ r represents growth rate

else, $b= 1-r$ r represents decay rate

Since b < 0.97:

$0.97= 1-r$

$r = 1 - 0.97$

$r = 0.03$

$r = 3\%$

$r = 0.03$

7 0

2 years ago

In Pennsylvania the average IQ score is 101.5. The variable is normally distributed, and the population standard deviation is 15

blsea [12.9K]

Answer:

We conclude that the Pennsylvania school district have an IQ higher than the average of 101.5

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 101.5

Sample mean, $\bar{x}$ = 106.4

Sample size, n = 30

Alpha, α = 0.05

Population standard deviation, σ = 15

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 101.5\\H_A: \mu > 101.5$

We use one-tailed z test to perform this hypothesis.

Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$z_{stat} = \displaystyle\frac{106.4 - 101.5}{\frac{15}{\sqrt{30}} } = 1.789$

Now, $z_{critical} \text{ at 0.05 level of significance } = 1.96$

Since,

$z_{stat} > z_{critical}$

We reject the null hypothesis and accept the alternate hypothesis.

Thus, we conclude that the Pennsylvania school district have an IQ higher than the average of 101.5

5 0

3 years ago