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2 years ago

Write an algebraic expression for the statement.The income earned at $ 10.91 per hour for w hours....

1 answer:

5 0

This is a "rate of pay" problem. The amount earned is equal to the (rate of pay) times the (number of hours worked).

Let the income be represented by "i". Then the formula is i = ($10.91/hour)*w, where w is the number of hours worked and has the unit of measurement "hours."

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The answer is red show step by step of the pro tem to get the answer

miskamm [114]

In order to rationalize the denominator of each expression, we need to multiply the expression by the same radical in the denominator, this way we can remove the radical from the denominator.

$\frac{5\sqrt{4}}{\sqrt{3}}(\cdot\sqrt{3})=\frac{5\sqrt{4}\sqrt{3}}{(\sqrt{3})^2}=\frac{5\cdot2\cdot\sqrt{3}}{3}=\frac{10\sqrt{3}}{3}$

10)

$-\frac{5}{3\sqrt{2}}(\cdot\sqrt{2})=-\frac{5\sqrt{2}}{3(\sqrt{2})^2}=-\frac{5\sqrt{2}}{3\cdot2}=-\frac{5\sqrt{2}}{6}$

11)

$\frac{2\sqrt{3}}{4\sqrt{5}}(\cdot\sqrt{5})=\frac{2\sqrt{3}\sqrt{5}}{4(\sqrt{5})^2}=\frac{2\sqrt{15}}{4\cdot5}=\frac{\sqrt{15}}{2\cdot5}=\frac{\sqrt{15}}{10}$

12)

$\frac{\sqrt{5}}{\sqrt{2}}(\cdot\sqrt{2})=\frac{\sqrt{5}\sqrt{2}}{(\sqrt{2})^2}=\frac{\sqrt{10}}{2}$

3 0

9 months ago

-7/12 + 3/8 = _____<br> help please.

In-s [12.5K]

Answer:

-5/24

Step-by-step explanation:

Find a common denominator:

-7/12 times 2= -14/24

3/8 times 3= 9/24

-14/24 +9/24 =5/24

8 0

2 years ago

Read 2 more answers

What is 1+1 pls I need help. LOL

Maru [420]

Answer:

.........................

༼ つ ◕_◕ ༽つ 2

8 0

2 years ago

411,500 science notation

horrorfan [7]

Answer:

the answer is 4.115 x 10^5

Step-by-step explanation:

hope that helps

7 0

3 years ago

We wish to see if the dial indicating the oven temperature for a certain model of oven is properly calibrated. Four ovens of thi

antoniya [11.8K]

Answer:

a. Standard deviation: 4.082

Standard error: 2.041

b. The 95% confidence interval for the actual temperature is (298.5, 311.5).

Upper bound: 311.5

Lower bound: 298.5

c. Test statistic t=2.45

P-value = 0.092

d. There is no enough evidence to claim that the dial of the oven is not properly calibrated. The actual temperature does not significantly differ from 300 °F.

e. If we use a significance level of 10% (a less rigorous test, in which the null hypothesis is rejected with with less requirements), the conclusion changes and now there is enough evidence to claim that the dial is not properly calibrated.

This happens because now the P-value (0.092) is smaller than the significance level (0.10), given statististical evidence for the claim.

Step-by-step explanation:

The mean and standard deviation of the sample are:

$M=\dfrac{1}{4}\sum_{i=1}^{4}(305+310+300+305)\\\\\\ M=\dfrac{1220}{4}=305$

$s=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{4}(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{3}\cdot [(305-(305))^2+(310-(305))^2+(300-(305))^2+(305-(305))^2]}\\\\\\ s=\sqrt{\dfrac{1}{3}\cdot [(0)+(25)+(25)+(0)]}\\\\\\ s=\sqrt{\dfrac{50}{3}}=\sqrt{16.667}\\\\\\s=4.082$

We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=305.

The sample size is N=4.

When σ is not known, s divided by the square root of N is used as an estimate of σM (standard error):

$s_M=\dfrac{s}{\sqrt{N}}=\dfrac{4.082}{\sqrt{4}}=\dfrac{4.082}{2}=2.041$

The degrees of freedom for this sample size are:

$df=n-1=4-1=3$

The t-value for a 95% confidence interval and 3 degrees of freedom is t=3.18.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=3.18 \cdot 2.041=6.5$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 305-6.5=298.5\\\\UL=M+t \cdot s_M = 305+6.5=311.5$

The 95% confidence interval for the actual temperature is (298.5, 311.5).

This is a hypothesis test for the population mean.

The claim is that the actual temperature of the oven when the dial is at 300 °F does not significantly differ from 300 °F.

Then, the null and alternative hypothesis are:

$H_0: \mu=300\\\\H_a:\mu\neq 300$

The significance level is 0.05.

The sample has a size n=4.

The sample mean is M=305.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=4.028.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{4.082}{\sqrt{4}}=2.041$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{305-300}{2.041}=\dfrac{5}{2.041}=2.45$

The degrees of freedom for this sample size are:

$df=n-1=4-1=3$

This test is a two-tailed test, with 3 degrees of freedom and t=2.45, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=2\cdot P(t>2.45)=0.092$

As the P-value (0.092) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the actual temperature of the oven when the dial is at 300 °F does not significantly differ from 300 °F.

If the significance level is 10%, the P-value (0.092) is smaller than the significance level (0.1) and the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the actual temperature of the oven when the dial is at 300 °C does not significantly differ from 300 °C.

5 0

2 years ago