All categories

2 years ago

Heellllppppppppppppp

Mathematics

2 answers:

daser333 [38]2 years ago

8 0

Answer:

x = $\frac{1}{5}$ , y = 6

Step-by-step explanation:

5x + y = 7 → (1)

20x + 2 = y → (2)

Substitute y = 20x + 2 into (1)

5x + 20x + 2 = 7

25x + 2 = 7 ( subtract 2 from both sides )

25x = 5 ( divide both sides by 25 )

x = $\frac{5}{25}$ = $\frac{1}{5}$

Substitute this value into (2) and evaluate

20( $\frac{1}{5}$ ) + 2 = y

4 + 2 = y , so

y = 6

marin [14]2 years ago

8 0

Answer:

o_O wut the heck did I just witness here?

Step-by-step explanation:

You might be interested in

Given WXYZ what is the measure of Z

nekit [7.7K]

Answer: 35

Step-by-step explanation:

5 0

3 years ago

Solve square root <br> (2x+3)^2=36

REY [17]

X=3/2 btw its a fraction

6 0

3 years ago

Read 2 more answers

140 is what percent of 500?

MAXImum [283]

Answer:

28%

Step-by-step explanation:

5 0

3 years ago

This is the whole picture can someone help??

Maksim231197 [3]

Step-by-step explanation:

I can't see good the picture

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

3 years ago