Y=2x+9 y=x+6 using substitution

The owner of a small store buys coats for $60.00 each. she sells the coats for $ 72.00 each. What percent of the purchase pric

tekilochka [14]

Answer:

Answer: 55% Step-by-step explanation: To find the percentage of the purchase price, simply divide the purchase price by the selling price and multiply it by 100. 50 90 0.55 0.55 * 100 = 55 This information is helpful to find a metric between supply and demand.

3 0

2 years ago

Please help. 21 points

Eduardwww [97]

Answer:

4/5 ÷3/5

4/5×5/3(any time one is dividing a fraction, the dominator would go up while the numerator would come down).

=4/3

4 0

2 years ago

A shop in the mall is 8 yards by 12 yards. How much would it cost to recarpet the shop with carpet that costs $2.69 per square y

valentinak56 [21]

Answer:

Step-by-step explanation:

You will first have to find the area to find the price of each square yard.

Area is 8*12 = 96. Multiply this by 2.69.

It will cost 258.24 dollars to recarpet the shop.

Let me know if this helped

7 0

1 year ago

Read 2 more answers

Camison's minnan gets 14 miles per gallon for city driving and 19 miles per gan

loris [4]

Answer:

7 gallons were used for city driving and 17 gallons were used for highway driving

Correct statement and question:

camison's minnan gets 14 miles per gallon for city driving and 19 miles per gallon for highway driving. At the beginning of the week, the 24-gallon tank was full. The family traveled 421 miles before running out of gas. How many gallons were used for city driving and how many were used for highway driving?

Source:

Previous question found at brainly

Step-by-step explanation:

x = Gallons for city driving

24 - x = Gallons for highway driving

Miles drove before running out of gas = 421

Now let's solve for x, writing the following equation:

14x + 19 (24 - x) = 421

14x + 456 - 19x = 421

-5x = 421 - 456

-5x = - 35

x = -35/-5

x = 7 ⇒ 24 -x = 17

7 gallons were used for city driving and 17 gallons were used for highway driving

3 0

2 years ago

Many high school students take the AP tests in different subject areas. In 2007, of the 144,796 students who took the biology ex

Nataly [62]

Answer:

$(0.582-0.485) - 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.0942$

$(0.582-0.485) + 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.09978$

And the 90% confidence interval would be given (0.0942;0.09978).

We are confident at 90% that the difference between the two proportions is between $0.0942 \leq p_A -p_B \leq 0.09978$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$p_A$ represent the real population proportion female for Biology

$\hat p_A =\frac{84199}{144796}=0.582$ represent the estimated proportion female for biology

$n_A=144796$ is the sample size for A

$p_B$ represent the real population proportion female for calculus AB

$\hat p_B =\frac{102598}{211693}=0.485$ represent the estimated proportion female for Calculus AB

$n_B=211693$ is the sample size required for B

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

For the 90% confidence interval the value of $\alpha=1-0.90=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.64$

And replacing into the confidence interval formula we got:

$(0.582-0.485) - 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.0942$

$(0.582-0.485) + 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.09978$

And the 90% confidence interval would be given (0.0942;0.09978).

We are confident at 90% that the difference between the two proportions is between $0.0942 \leq p_A -p_B \leq 0.09978$

5 0

2 years ago