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3 years ago

PLEASE HELP ASAP please show how you got the answer

Mathematics

1 answer:

Mashcka [7]3 years ago

4 0

Answer:

no wonder you're posting this here, there isn't enough information to solve this problem. I would just screenshot this to your teacher or something.

sorry it took so long to respond, my last answer got deleted for some reason :/

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Are the linear equations in standard form? sort each equation into the appropriate category.

Sholpan [36]

What equation? You would have to pass a photo, but...
Standard form is: ax+by=c
Slope intercept form is: y=my+b
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3 years ago

In science class, the students observed 24 of the 300 frogs from the local aquarium. If 6 of the frogs observed had spots, how m

Alona [7]

Answer:

225 frogs

Step-by-step explanation:

Total population of frogs = 300 frogs.

Observed population of frogs = 24

6 of the 24 observed frogs had spots

Which means , the number of frogs that did not have spots = 24 - 6 = 18 frogs.

We were told to find how many of the total population can be predicted to NOT have spots. We would form a proportion.

If 24 frogs = 18 frogs with no spots

300 frogs = Y

Cross multiply

24Y = 300 × 18

Y = (300 × 18) ÷ 24

Y = 5400 ÷ 24

Y = 225 frogs.

This means out of 300 frogs, 225 frogs do not have spots.

Therefore, the total population that can be predicted to NOT have spots is 225 frogs.

the total population can be predicted to NOT have spots

3 0

3 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.