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3 years ago

How many milliliters of water are in a 2- liter bottle of water

Mathematics

1 answer:

e-lub [12.9K]3 years ago

6 0

There are 1000 milliliters per liter. So there are 2,000 milliliters in 2 liters

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Choice 1: Angela and Kellie went to lunch. Their bill before tax was $20.00. Tax is 8%. How much tax should

Zolol [24]

Answer:

$1.60

Step-by-step explanation:

Amount of tax paid = percentage of tax to be paid x bill before tax

0.08 x $20 = $1.60

4 0

3 years ago

Can someone help me please

Nitella [24]

Answer:

16 units

Step-by-step explanation:

The perimeter is the sum of the lengths

Add up the lengths as shown

3+2+1+2+3+1+1+3

16 units

4 0

3 years ago

Draw a line representing the rise And a line representing the run of the mind state is slope of the line in simplest form

mamaluj [8]

Answer:

The slope is (1/5)

Step-by-step explanation:

See attached

7 0

2 years ago

In how many years will the interest of a sum of rs 3600 at the rate of 5.5% per year be rs 594?

boyakko [2]

Answer:

In 3 years, the interest of a sum of rs 3600 at the rate of 5.5% per year be Rs 594.

Step-by-step explanation:

T = ?

R = 5.5%

P = Rs 3600

I = Rs 594

T = (I×100)/(P×R)

or, T = (594×100)/(3600×5.5)

or, T = 59400/19800

so, T = 3 yrs

7 0

3 years ago

The USA Today reports that the average expenditure on Valentine's Day is $100.89. Do male and female consumers differ in the amo

Aloiza [94]

Answer:

$a)\ \ \bar x_m-\bar x_f=67.03\\\\b)\ \ E=15.7416\\\\c)\ \ CI=[51.2884, \ 82.7716]$

Step-by-step explanation:

a. -Given that:

$n_m=41\ , \ \sigma_m=33, \bar x_m=135.67\\\\n_f=37. \ \ ,\sigma_f=20, \ \ \bar x_f=68.64$

#The point estimator of the difference between the population mean expenditure for males and the population mean expenditure for females is calculated as:

$\bar x_m-\bar x_f\\\\\therefore \bigtriangleup\bar x=135.67-68.64\\\\=67.03$

Hence, the pointer is estimator 67.03

b. The standard error of the point estimator, $\bar x_m-\bar x_f$ is calculated by the following following:

$\sigma_{\bar x_m-\bar x_f}=\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}$

-And the margin of error, E at a 99% confidence can be calculated as:

$E=z_{\alpha/2}\times \sigma_{\bar x_m-\bar x_f}\\\\\\=z_{0.005}\times\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}\\\\\\=2.575\times \sqrt{\frac{33^2}{41}+\frac{20^2}{37}}\\\\\\=15.7416$

Hence, the margin of error is 15.7416

c. The estimator confidence interval is calculated using the following formula:

$\bar x_m-\bar x_f\ \pm z_{\alpha/2}\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}$

#We substitute to solve for the confidence interval using the standard deviation and sample size values in a above:

$CI=\bar x_m-\bar x_f\ \pm z_{\alpha/2}\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}\\\\=(135.67-68.64)\pm 15.7416\\\\=67.03\pm 15.7416\\\\=[51.2884, \ 82.7716]$

Hence, the 99% confidence interval is [51.2884,82.7716]

7 0

3 years ago