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3 years ago

Are the ratios 2:3 and 10:18 equivalentYES OR NO

Mathematics

2 answers:

netineya [11]3 years ago

6 0

Yes they are equivalent

VashaNatasha [74]3 years ago

4 0

Answer:

Step-by-step explanation:

an equivalent ration to 2:3 would be 10:15. because when you simpliy 10:15 the most you can you would get 2:3. where as 10:18 is not equal

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morpeh [17]

If we solve this equation: 8.9=logx+7.9

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5 0

3 years ago

A rectangular parking lot must have a perimeter of 440 feet and an area of at least 8000 square feet. Describe the possible leng

tangare [24]

Answer:

The possible parking lengths are 45.96 feet and 174.031 feet

Step-by-step explanation:

Let x be the length of rectangular plot and y be the breadth of rectangular plot

A rectangular parking lot must have a perimeter of 440 feet

Perimeter of rectangular plot =2(l+b)=2(x+y)=440

2(x+y)=440

x+y=220

y=220-x

We are also given that an area of at least 8000 square feet.

So, $xy \leq 8000$

So, $x(220-x) \leq 8000$

$220x-x^2 \leq 8000$

So, $220x-x^2 = 8000\\-x^2+220x-8000=0$

General quadratic equation : $ax^2+bx+c=0$

Formula : $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-220 \pm \sqrt{220^2-4(-1)(-8000)}}{2(-1)}\\x=\frac{-220 + \sqrt{220^2-4(-1)(-8000)}}{2(-1)} , \frac{-220 - \sqrt{220^2-4(-1)(-8000)}}{2(-1)}\\x=45.96,174.031$

So, The possible parking lengths are 45.96 feet and 174.031 feet

3 0

3 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

Please help me with part (iii). If possible help me check my answers too. Thanks

zhannawk [14.2K]

(i) There is glucose outside the Visking tubing. Glucose is small enough to pass through it.

(iii) Normal urine color ranges from pale yellow to deep amber — the result of a pigment called urochrome and how diluted or concentrated the urine is. Pigments and other compounds in certain foods and medications can change your urine color. Beets, berries and fava beans are among the foods most likely to affect the color.

LINK: https://www.mayoclinic.org/diseases-conditions/urine-color/symptoms-causes/syc-20367333

I HOPE THIS HELPS. IF IT DOES, PLEASE MARK BE AS BRAINLIEST. EVERYTHING ELSE IS CORRECT EXEPT (i)

5 0

3 years ago

In circle M, diameters JL and HK each measure 16

creativ13 [48]

Answer:

3.5 cm

Step-by-step explanation:

In the picture attached, circle M is shown.

From the picture, minor arc JH measures 25°.

The length of the circle is π*Diameter = π*16 = 50.26 cm. This length is proportional to 360°. To find the length of minor arc JH, we have to use the next proportion:

50.26 cm / x cm = 360° / 25°

x = 50.26*25/360

x = 3.5 cm

5 0

3 years ago

Are the ratios 2:3 and 10:18 equivalentYES OR NO​

Are the ratios 2:3 and 10:18 equivalentYES OR NO