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3 years ago

(picture) is this the correct way to solve 19?

Mathematics

1 answer:

goldfiish [28.3K]3 years ago

8 0

I cant see your work very clearly so I'll work it out below:-

f(x + Δx) = (x + Δx)^2 - 2(x + Δx) - 3

= x^2 + 2x(Δx) + (Δx)^2 - 2x - 2(Δx) - 3

= x^2 - 2x + (Δx)^2 + 2x(Δx) - 2(Δx) - 3

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The first discount on a camera was 18%. The second discount was 20%. After these two discounts the price was $328. What was the

Artyom0805 [142]

Answer:

$500

Step-by-step explanation:

We can find the original price of the camera through a proportion. A proportion is an equation where two ratios or fractions are equal. The ratios or fractions compare like quantities.

<u>Second Discount</u>

20% off means we paid 80%. We know we paid $328 of some price.

$\frac{80}{100} =\frac{328}{y}$

I can now cross-multiply by multiplying numerator and denominator from each ratio.

$80y=100(328)\\80y=32800$

I now solve for y by dividing by 80.

$\frac{80y}{80} =\frac{32800}{80} \\y= $410$

The price after the first discount was $410.

<u>First Discount</u>

We will repeat the steps above with $410. 18% off means we paid 82%.

$\frac{82}{100} =\frac{410}{y}$

I can now cross-multiply by multiplying numerator and denominator from each ratio.

$82y=100(410)\\82y=41000$

I now solve for y by dividing by 82.

$\frac{82y}{82} =\frac{41000}{82} \\y= $500$

The original price was $500.

5 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:

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