Find x. (More info in pic)

Explain how the model represents 5/ 1third

Ilia_Sergeevich [38]

Here is ur perfect answer

3 0

2 years ago

The product of a number and -7 is at most -70. what is inequality 18 points

zimovet [89]

Answer:

Inequality :

$-7x\le \:-70$

Step-by-step explanation:

Let the unknown number be x

at most -70 = ≤ -70

$x\times(-7) \leq -70\\\\-7x\leq -70$

Solving further :

$-7x\le \:-70\\\\\mathrm{Multiply\:both\:sides\:by\:-1\:\left(reverse\:the\:inequality\right)}\\\\\left(-7x\right)\left(-1\right)\ge \left(-70\right)\left(-1\right)\\\\Simplify\\\\7x\geq 70\\\\\mathrm{Divide\:both\:sides\:by\:}7\\\\\frac{7x}{7}\ge \frac{70}{7}\\\\Simplify\\\\x \geq 10$

3 0

2 years ago

Find the value of a/b/c if a=1/3, b=1/2, c=1/6

densk [106]

Answer: The answer is 0.1111111111111

Step-by-step explanation:

8 0

3 years ago

The mean annual premium for automobile insurance in the United States is $1503 (Insure website, March 6, 2014). Being from Penns

romanna [79]

Answer:

If we compare the p value and the significance level for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the mean annual premium in Pennsylvania is significantly lower than the national mean annual premium of 1503 at 5% of significance.

Step-by-step explanation:

1) Data given and notation

$\bar X=1440$ represent the mean annual premium value for the sample

$s=165$ represent the sample standard deviation for the sample

$n=25$ sample size

$\mu_o =1503$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean annual premium in Pennsylvania is lower than the national mean annual premium, the system of hypothesis would be:

Null hypothesis: $\mu \geq 1503$

Alternative hypothesis: $\mu < 1503$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{1440-1503}{\frac{165}{\sqrt{25}}}=-1.909$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=25-1=24$

Since is a one side left tailed test the p value would be:

$p_v =P(t_{(24)}$

Conclusion

If we compare the p value and the significance level for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the mean annual premium in Pennsylvania is significantly lower than the national mean annual premium of 1503 at 5% of significance.

7 0

3 years ago

Five carpenters worked 15 & 3/4, 15 & 1/4, 17 & 1/2, 12 & 1/2 and 17 & 1/2 hours respectively. What was the

azamat

<u>The total </u><u>labor cost</u><u> of the carpenters is $2296.91</u>

The carpenters worked $15\frac{3}{4} , 15\frac{1}{4}, 17\frac{1}{2}, 12\frac{1}{2}, 17\frac{1}{2}$