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2 years ago

PLEASE HELP!!!!! PEOPLE ARE GETTING THEM WRONG FOR ME!!!

Mathematics

1 answer:

UkoKoshka [18]2 years ago

8 0

Answer:

18 lawns

Step-by-step explanation:

Lets take the number of lands that would be mow(ed) in 36 hours as x

Hence,

$Thriugh\ the\ proportion\ method\ , we\ get\ \\\\3 : 6::x:36\\36*3=6x\\\frac{36*3}{6} =x\\\frac{108}{6}=x\\x=18$

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Jeffery is playing a game with two spinners below:

Mrrafil [7]

The <em>correct answer</em> is:

{3, 4, 5, 6, 7, 8, 9, 10, 11, 12} .

Explanation:

The sample space for rolling these two spinners is:

(1, 2), (1, 4), (1, 6), (1, 8)

(2, 2), (2, 4), (2, 6), (2, 8)

(3, 2), (3, 4), (3, 6), (3, 8)

(4, 2), (4, 4), (4, 6), (4, 8)

This gives us the sums:

3, 5, 7, 9

4, 6, 8, 10

5, 7, 9, 11

6, 8, 10, 12

As a set, these are:

{3, 4, 5, 6, 7, 8, 9, 10, 11 12}

3 0

3 years ago

Read 2 more answers

Beach Cruises prints brochures and fliers to advertise their dolphin watching tours. To print, the brochure requires three pages

Nimfa-mama [501]

Answer:

500 divided by 80 equals 6.25. 6.25 divided by 100 equals 0.0625

Step-by-step explanation:

500 divided by 80 equals 6.25. 6.25 divided by 100 equals 0.0625

6 0

3 years ago

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

Holly bought blank CD's and spent $34.57.What is that amount rounded to the nearest dollar?.

Tamiku [17]

Answer:

$35

Step-by-step explanation:

Rounding to the nearest dollar which would be 4 so in that case you would use the 5 in .57 and because 5 is a large number it would make the 4 into a five thus making it $35. Hope this helped :)

3 0

2 years ago

Find x and y, given that line WS and line VT are parallel. Show all work!

bulgar [2K]

In the given diagram, the traingles USW and UTV are similar triangles and thus the following ratio equality applies to them.

$\frac{VT}{WS} =\frac{VU}{WU}=\frac{TU}{SU}$ ..........(Equation 1)