Ursula hiked 69 kilometers through<br> the woods. How many decimeters did<br> she hike?

Michael is 12 yrs older than Ava. The sum of their ages are 38. What are their ages?

Setler [38]

She is 26 years old because you need to do 38 - 12=26!

5 0

3 years ago

Which of the following is the additive inverse of 4/31 ?

zhuklara [117]

Answer:

c 31b4

Step-by-step explanation:

5 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

4 years ago

Farmers know that driving heavy equipment on wet soil compresses the soil and injures future crops. Here are data on the "penetr

Greeley [361]

Answer:

Step-by-step explanation:

Hello!

To see if driving heavy equipment on wet soil compresses it causing harm to future crops, the penetrability of two types of soil were measured:

Sample 1: Compressed soil

X₁: penetrability of a plot with compressed soil.

n₁= 20 plots

X[bar]₁= 2.90

S₁= 0.14

Sample 2: Intermediate soil

X₂: penetrability of a plot with intermediate soil.

n₂= 20 (with outlier) n₂= 19 plots (without outlier)

X[bar]₂= 3.34 (with outlier) X[bar]₂= 2.29 (without outlier)

S₂= 0.32 (with outlier) S₂= 0.24 (without outlier)

Outlier: 4.26

Assuming all conditions are met and ignoring the outlier in the second sample, you have to construct a 99% CI for the difference between the average penetration in the compressed soil and the intermediate soil. To do so, you have to use a t-statistic for two independent samples:

Parámeter of interest: μ₁-μ₂

Interval:

[(X[bar]₁-X[bar]₂)± $t_{n_1+n_2-2;1-\alpha/2}$ *Sa $\sqrt{\frac{1}{n_1} +\frac{1}{n_2} }$ ]