Help me with this please

A certificate of deposit earns 0.75% interest every three months. The interest is compounded. What is the value of a $32,000 inv

leonid [27]

480,000 i think this is the answer

7 0

3 years ago

Read 2 more answers

Bob is buying the soccer team snacks after their game today. 8 players want cookies and 5 players want ice cream cones. If ice c

kotykmax [81]

Answer:

$20.25

Step-by-step explanation:

8*1.75=x

x=14

5*1.25= y

y=6.25

x+y=20.25

3 0

2 years ago

Sara buys 3 of the same type of dress, but in different colors. If she spends a total of $80.34, how much does each dress cost?

Musya8 [376]

Answer:

80.34 / 3 = 26.78

5 0

3 years ago

Read 2 more answers

An accounting firm is planning for the next tax preparation season. From last years returns, the firm collects a systematic rand

Elena L [17]

Answer:

a)From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the standard error for the mean would be:

$\sigma_{\bar X}= \frac{140}{\sqrt{100}} =14$

b) We want this probability:

$P(\bar X >120)$

And we can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143$

And we can find this probability with the complement rule and the normal standard deviation or excel and we got:

$P( z>2.143) = 1-P(Z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

Part a

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the standard error for the mean would be:

$\sigma_{\bar X}= \frac{140}{\sqrt{100}} =14$

Part b

We want this probability:

$P(\bar X >120)$

And we can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143$

And we can find this probability with the complement rule and the normal standard deviation or excel and we got:

$P( z>2.143) = 1-P(Z$

4 0

3 years ago

3y + 2x - 10 = 0 <br>find m

Hitman42 [59]

Step-by-step explanation:

3y + 2x - 10 = 0

in standard form

2x + 3y -10=0

||

as (ax + by - c = 0)

m = -a/b

= -2/3

or

3y + 2x - 10 = 0

make y subject

y = -2/3x + 10/3

6 0

2 years ago