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2 years ago

Between them Molly and Sue have 32 marbles. Molly has 3 times as many as Sue. How many marbles does each girl have?

Mathematics

1 answer:

I am Lyosha [343]2 years ago

3 0

Answer:

Sue has 8 marbles

Molly has 8 marbles

Step-by-step explanation:

m = number of marbles Molly has

s = number of marbles that sue has

m+s =32

m = 3s

Substitute the second equation into the first equation

3s+s =32

4s = 32

Divide by 4

4s/4 =32/4

s = 8

Sue has 8 marbles

m = 32 = 3*8 = 24

Molly has 8 marbles

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Answer:

Step-by-step explanation:

There are 2 yellow out of 10.

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3 years ago

Tanner and Zach both play basketball. In a game, Tanner scored six points. Zach scored four times as many times as Tanner. How m

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Answer:

Step-by-step explanation:

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3 years ago

Please help me on this!! 5n+17≥n–3≥–15+5n

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Answer:

-5 ≤ n ≤ 3

Step-by-step explanation:

Firstly we solve the equation two at a time;

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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.

$z_{\alpha/2}=1.64$

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