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1 year ago

3. Last month, Ronald sold 52 packages

Mathematics

1 answer:

vladimir2022 [97]1 year ago

8 0

Answer:

$468.75

Step-by-step explanation:

52 packages = $195

125 packages = ?

125 x (195/52)

= 125 x 3.75

= 468.75

You might be interested in

Enter your answers in the boxes to correctly complete this statement.

harina [27]

Answer:

349800

Step-by-step explanation:

First Part:

The question tells you to multiply by 10 to the power of 4.

This implies that you can move the decimal point 4 numbers to the right

When you move the decimal two numbers to the right you get 3498

Then you got to move it two more numbers to the right

3498 can also be written as 3498.0000000000 (infinite amount of zeros)

So moving the decimal two more places you get 349800

Second Part:

10^4 = 10 * 10 * 10 * 10

= 10000

Count the zeros you get 4.

3 0

2 years ago

An article reported that for a sample of 52 kitchens with gas cooking appliances monitored during a one-week period, the sample

kolbaska11 [484]

Answer:

a) $654.16-2.01\frac{164.55}{\sqrt{52}}=608.29$

$654.16+2.01\frac{164.55}{\sqrt{52}}=700.03$

And we can conclude that we are 95% confident that the true mean of Co2 level is between 608.29 and 700.03 ppm

b) $n=(\frac{1.960(175)}{25})^2 =188.23 \approx 189$

Step-by-step explanation:

Part a

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:

$df=n-1=52-1=51$

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 table to find the critical value. The excel command would be: "=-T.INV(0.025,51)".And we see that $t_{\alpha/2}=2.01$

Replacing we got:

$654.16-2.01\frac{164.55}{\sqrt{52}}=608.29$

$654.16+2.01\frac{164.55}{\sqrt{52}}=700.03$

And we can conclude that we are 95% confident that the true mean of Co2 level is between 608.29 and 700.03 ppm

Part b

The margin of error is given by :

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

The desired margin of error is ME =50/2=25 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} s}{ME})^2$ (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got $z_{\alpha/2}=1.960$ , and we use an estimator of the population variance the value of 175 replacing into formula (b) we got: