W is greater than -5 and less than 0

A man invests his savings in two accounts ,one paying6% and the other paying 10% simple interest per year. He puts twice as much

AnnyKZ [126]

Answer: he invested $46062.5 at 6% and $23031.25 at 10%

Step-by-step explanation:

Let x represent the amount which he invested in the account paying 6% interest.

Let y represent the amount which he invested in the account paying 10% interest.

He puts twice as much in the lower-yielding account because it is less risky.. This means that

x = 2y

The formula for determining simple interest is expressed as

I = PRT/100

Considering the account paying 6% interest,

P = $x

T = 1 year

R = 6℅

I = (x × 6 × 1)/100 = 0.06x

Considering the account paying 10% interest,

P = $y

T = 1 year

R = 10℅

I = (y × 10 × 1)/100 = 0.1y

His annual interest is $7370dollars. it means that

0.06x + 0.2y = 7370 - - - - - - - - - -1

Substituting x = 2y into equation 1, it becomes

0.06 × 2y + 0.2y = 7370

0.12y + 0.2y = 7370

0.32y = 7370

y = 7370/0.32

y = $23031.25

x = 2 × 23031.25

x = 46062.5

6 0

3 years ago

Write (-3)^4 in standard form

Olenka [21]

Answer: 81

Step-by-step explanation: Because, (-3)^4=-3*-3*-3*-3=-(-81)=81

3 0

3 years ago

The 2008 Workplace Productivity Survey, commissioned by LexisNexis and prepared by WorldOne Research, included the question, "Ho

vitfil [10]

Answer:

Therefore, the sampling distribution of $\bar{x}$ is normal with a mean equal to 9 hours and a standard deviation of 0.7969 hours.

The 95% interval estimate of the population mean $\mu$ is

LCL = 7.431 hours to UCL = 10.569 hours

Step-by-step explanation:

Let X be the number of hours a legal professional works on a typical workday. Imagine that X is normally distributed with a known standard deviation of 12.6.

The population standard deviation is

$\sigma = 12.6 \: hours$

A sample of 250 legal professionals was surveyed, and the sample's mean response was 9 hours.

The sample size is

$n = 250$

The sample mean is

$\bar{x} = 9 \: hours$

Since the sample size is quite large then according to the central limit theorem, the sample mean is approximately normally distributed.

The population mean would be the same as the sample mean that is

$\mu = \bar{x} = 9 \: hours$

The sample standard deviation would be

$s = {\frac{\sigma}{\sqrt{n} }$

Where is the population standard deviation and n is the sample size.

$s = {\frac{12.6}{\sqrt{250} }$

$s = 0.7969 \: hours$

Therefore, the sampling distribution of $\bar{x}$ is normal with a mean equal to 9 hours and a standard deviation of 0.7969 hours.

The population mean confidence interval is given by

$\text {confidence interval} = \mu \pm MoE\\\\$

Where the margin of error is given by

$$ MoE = t_{\alpha/2}(\frac{s}{\sqrt{n} } ) $ \\\\$

Where n is the sampling size, s is the sample standard deviation and is the t-score corresponding to a 95% confidence level.

The t-score corresponding to a 95% confidence level is

Significance level = α = 1 - 0.95 = 0.05/2 = 0.025

Degree of freedom = n - 1 = 250 - 1 = 249

From the t-table at α = 0.025 and DoF = 249

t-score = 1.9695

$MoE = t_{\alpha/2}(\frac{\sigma}{\sqrt{n} } ) \\\\MoE = 1.9695\cdot \frac{12.6}{\sqrt{250} } \\\\MoE = 1.9695\cdot 0.7969\\\\MoE = 1.569\\\\$

So the required 95% confidence interval is

$\text {confidence interval} = \mu \pm MoE\\\\\text {confidence interval} = 9 \pm 1.569\\\\\text {LCI } = 9 - 1.569 = 7.431\\\\\text {UCI } = 9 + 1.569 = 10.569$