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3 years ago

6

The minimum wage in 2003 was $5.15. This was W more than the minimum wage in 1996. Write an expression for the minimum wage in 1

996.

1 answer:

allsm [11]3 years ago

6 0

Look it up again on the computer

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Simplify y+1<3 please need help

Zepler [3.9K]

STEP-BY-STEP SOLUTION:

Let's first establish the inequality which we need to solve as displayed below:

y + 1 < 3

To begin with, we need to make ( y ) the subject by keeping it on the left-hand side of the inequality and placing all other numbers on the right-hand side of the equality as displayed below:

y + 1 < 3

y < 3 - 1

Then, we simply simplify / solve as displayed below:

y < 3 - 1

y < 2

ANSWER:

y < 2

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3 years ago

Solve the following proportion for y.

LekaFEV [45]

Answer:

y = 110/3 or 36.6 (the 6 is infinite so make a line over it when entering the answer)

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3 years ago

Can somebody help me wit this ? Cause Ion understand this .

Triss [41]

Answer:

wow that's nice of your

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3 years ago

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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 $0.0942 \leq p_A -p_B \leq 0.09978$

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3 years ago

The temperature in Alaska can reach -30 degrees Fahrenheit during the winter months it is colder in Antarctica than in the Alask

Fofino [41]

Answer:

B. -40°F

Step-by-step explanation:

its- clearly lower than -30

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3 years ago

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