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2 years ago

7

Leah has 2/5 of a gallon of paint. She decides to use 1/4 of this paint to paint a door. What fraction of a gallon of paint does

she use for the door? Explain your answer.

1 answer:

love history [14]2 years ago

3 0

Answer:

$\frac{1}{10}$

Step-by-step explanation:

1/4 of 2/5 = 1/4 * 2/5

= $\frac{2}{20}$

$\frac{2}{20}$ could be simplified into $\frac{1}{10}$

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A grower believes that one in five of his citrus trees are infected with the citrus red mite. How large a sample should be taken

mihalych1998 [28]

Answer:

n=6147

Step-by-step explanation:

1) Notation and definitions

$X=1$ number of citrus trees that are infected with the citrus red mite.

$n=5$ random sample taken

$\hat p=\frac{1}{5}=0.2$ estimated proportion of citrus trees that are infected with the citrus red mite.

$p$ true population proportion of citrus trees that are infected with the citrus red mite.

Me=0.01 represent the margin of error desired

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".

The population proportion have the following distribution

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

In order to find the critical values we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical values would be given by:

$t_{\alpha/2}=-1.96, t_{1-\alpha/2}=1.96$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.01$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.2(1-0.2)}{(\frac{0.01}{1.96})^2}=6146.56$

And rounded up we have that n=6147

5 0

3 years ago

Find the sum of the first 50 positive even integers.

adell [148]

2250
50*(2+100)/2
50*102/2
50*51
2250

7 0

3 years ago

A department store sells a pair of shoes with a 80% markup. If the store bought the pair of shoes for $55.25. Round the price to

AleksandrR [38]

Answer:

$31.00

Step-by-step explanation:

that is the procedure above

6 0

3 years ago

A couple intends to have two children, and suppose that approximately 52% of births are male and 48% are female.

Pachacha [2.7K]

a) Probability of both being males is 27%

b) Probability of both being females is 23%

c) Probability of having exactly one male and one female is 50%

Step-by-step explanation:

a)

The probability that the birth is a male can be written as

$p(m) = 0.52$ (which corresponds to 52%)

While the probability that the birth is a female can be written as

$p(f) = 0.48$ (which corresponds to 48%)

Here we want to calculate the probability that over 2 births, both are male. Since the two births are two independent events (the probability of the 2nd to be a male does not depend on the fact that the 1st one is a male), then the probability of both being males is given by the product of the individual probabilities:

$p(mm)=p(m)\cdot p(m)$

And substituting, we find

$p(mm)=0.52\cdot 0.52 = 0.27$

So, 27%.

b)

In this case, we want to find the probability that both children are female, so the probability

$p(ff)$

As in the previous case, the probability of the 2nd child to be a female is independent from whether the 1st one is a male or a female: therefore, we can apply the rule for independent events, and this means that the probability that both children are females is the product of the individual probability of a child being a female:

$p(ff)=p(f)\cdot p(f)$

And substituting

$p(f)=0.48$

We find:

$p(ff)=0.48\cdot 0.48=0.23$

Which means 23%.

c)

In this case, we want to find the probability they have exactly one male and exactly one female child. This is given by the sum of two probabilities:

- The probability that 1st child is a male and 2nd child is a female, namely $p(mf)$

- The probability that 1st child is a female and 2nd child is a male, namely $p(fm)$

So, this probability is

$p(mf Ufm)=p(mf)+p(fm)$

We have:

$p(mf)=p(m)\cdot p(f)=0.52\cdot 0.48=0.25$

$p(fm)=p(f)\cdot p(m)=0.48\cdot 0.52=0.25$

Therefore, this probability is

$p(mfUfm)=0.25+0.25=0.50$

So, 50%.

Learn more about probabilities:

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5 0

3 years ago

PLEASE HELP! my test is due in 10 minutes.. What is teh range of the relation above?

lisov135 [29]

Answer:

D) The Y value never goes above 6.

Step-by-step explanation:

7 0

3 years ago