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2 years ago

BIG POINTS ANSWER WITH DETAIL! Solve for x: 3x + 1 = 3x

1 answer:

6 0

There are no solutions to this problem,

Let's solve your equation step-by-step.
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Step 1: Subtract 3x from both sides.
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Step 2: Subtract 1 from both sides.
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Megan uses 2/3 cups of almonds to make 4 cups of trail mix. Using the same ratio, how many cups of almonds would megan need to m

lakkis [162]

1 and 1/2
9 divided by 4 = 2.25
2.25x2/3=1 1/2

3 0

3 years ago

Read 2 more answers

The average summer temperature in Anchorage is 69°F. If the daily temperature is normally distributed with a standard deviation

mafiozo [28]

Answer:

81.85%

Step-by-step explanation:

Given :

The average summer temperature in Anchorage is 69°F.

The daily temperature is normally distributed with a standard deviation of 7°F .

To Find:What percentage of the time would the temperature be between 55°F and 76°F?

Solution:

Mean = $\mu = 69$

Standard deviation = $\sigma = 7$

Formula : $z=\frac{x-\mu}{\sigma}$

Now At x = 55

$z=\frac{55-69}{7}$

$z=-2$

At x = 76

$z=\frac{76-69}{7}$

$z=1$

Now to find P(55<z<76)

P(2<z<-1)=P(z<2)-P(z>-1)

Using z table :

P(2<z<-1)=P(z<2)-P(z>-1)=0.9772-0.1587=0.8185

Now percentage of the time would the temperature be between 55°F and 76°F = $0.8185 \times 100 = 81.85\%$

Hence If the daily temperature is normally distributed with a standard deviation of 7°F, 81.85% of the time would the temperature be between 55°F and 76°F.

7 0

2 years ago

An examination of the records for a random sample of 16 motor vehicles in a large fleet resulted in the sample mean operating co

levacccp [35]

Answer:

1. The 95% confidence interval would be given by (24.8190;27.8010)

2. $4.7048 \leq \sigma^2 \leq 16.1961$

3. $t=\frac{26.31-25}{\frac{2.8}{\sqrt{16}}}=1.871$

$t_{crit}=1.753$

Since our calculated value it's higher than the critical value we have enough evidence at 5% of significance to reject th null hypothesis and conclude that the true mean is higher than 25 cents per mile.

4. $t=(16-1) [\frac{2.8}{2.3}]^2 =22.2306$

$\chi^2 =24.9958$

Since our calculated value is less than the critical value we don't hav enough evidence to reject the null hypothesis at the significance level provided.

Step-by-step explanation:

Previous concepts

$\bar X=26.31$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=2.8 represent the sample standard deviation

n=16 represent the sample size

Part 1

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=16-1=15$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,15)".And we see that $t_{\alpha/2}=2.13$

Now we have everything in order to replace into formula (1):

$26.31-2.13\frac{2.8}{\sqrt{16}}=24.819$

$26.31+2.13\frac{2.8}{\sqrt{16}}=27.801$

So on this case the 95% confidence interval would be given by (24.8190;27.8010)

Part 2

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=16-1=15$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical values.

The excel commands would be: "=CHISQ.INV(0.05,15)" "=CHISQ.INV(0.95,15)". so for this case the critical values are:

$\chi^2_{\alpha/2}=24.996$

$\chi^2_{1- \alpha/2}=7.261$

And replacing into the formula for the interval we got:

$\frac{(15)(2.8)^2}{24.996} \leq \sigma^2 \leq \frac{(15)(2.8)^2}{7.261}$

$4.7048 \leq \sigma^2 \leq 16.1961$

Part 3

We need to conduct a hypothesis in order to determine if actual mean operating cost is at most 25 cents per mile , the system of hypothesis would be:

Null hypothesis: $\mu \leq 25$

Alternative hypothesis: $\mu > 25$

We don't know the population deviation, so for this case 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)

Calculate the statistic

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

$t=\frac{26.31-25}{\frac{2.8}{\sqrt{16}}}=1.871$

Critical value

On this case we need a critical value on th t distribution with 15 degrees of freedom that accumulates 0.05 of th area on the right and 0.95 of the area on the left. We can calculate this value with the following excel code:"=T.INV(0.95,15)" and we got $t_{crit}=1.753$

Conclusion

Since our calculated valu it's higher than the critical value we have enough evidence at 5% of significance to reject th null hypothesis and conclude that the true mean is higher than 25 cents per mile.

Part 4

State the null and alternative hypothesis

On this case we want to check if the population standard deviation is more than 2.3, so the system of hypothesis are:

H0: $\sigma \leq 2.3$

H1: $\sigma >2.3$

In order to check the hypothesis we need to calculate the statistic given by the following formula:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.

What is the value of your test statistic?

Now we have everything to replace into the formula for the statistic and we got:

$t=(16-1) [\frac{2.8}{2.3}]^2 =22.2306$

What is the critical value for the test statistic at an α = 0.05 significance level?

Since is a right tailed test the critical zone it's on the right tail of the distribution. On this case we need a quantile on the chi square distribution with 15 degrees of freedom that accumulates 0.05 of the area on the right tail and 0.95 on the left tail.

We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.95,15)". And our critical value would be $\chi^2 =24.9958$

Since our calculated value is less than the critical value we FAIL to reject the null hypothesis.

7 0

3 years ago

What is the gradient of the blue line????<br> help please

Maru [420]

Answer:

-5

Step-by-step explanation:

gradient = change in y ÷ change in x

gradient = -5 ÷ 1 = -5

8 0

2 years ago

4. A triangle with base (2x+2) and height (x-2) is cut out of a rectangular sheet of paper that is (3x+4) by (2x+3) what is the

juin [17]

So what we do is
area that remains=total area-triangle area that was cut out

we need to find 2 things
total area
triangle area

total area=rectange=base times height
area=(3x+4) times (2x+3)
FOIL or distribute
6x^2+8x+9x+12=6x^2+17x+12

triangle area=1/2 times base times height
triangle area=1/2 times (2x+2) times (x-2)=
(x+2) times (x-2)=x^2+2x-2x-4=x^2-4

so
total area=6x^2+17x+12
triangle area=x^2-4

subtract
area that remains=total area-triangle area that was cut out
area that remains=6x^2+17x+12-(x^2-4)=
6x^2+17x+12-x^2+4=
6x^2-x^2+17x+12+4=
5x^2+17x+16

area that remains is 5x^2+17x+16

5 0

2 years ago