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1 year ago

Bill notes that his apple tree is 4/5 the size of his pear tree is apple tree is 4 M tall how tall is the pear tree

Mathematics

1 answer:

e-lub [12.9K]1 year ago

7 0

1) Given that the apple tree is 4/5 pear tree we can write

Apple Tree size = 4/5 Peartree

Apple tree= 4m

2) So we can write the following expression for that:

$\begin{gathered} Apple=\frac{4}{5}P \\ 4=\frac{4}{5}_{}P \\ 4P=20m \\ P=\frac{20}{4} \\ P=5 \\ \\ \end{gathered}$

So the Pear tree is 5 meters tall, and the apple tree is 4/5*5 = 4 m

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2 years ago

Complete parts (a) through (c) below. (a) Determine the critical value(s) for a right-tailed test of a population mean at t

olga_2 [115]

Answer:

a) The critical value on this case would be $t_{crit}=1.325$

b) The critical value on this case would be $t_{crit}=-1.345$

c) The critical values on this case would be $t_{crit}=\pm 2.201$

Step-by-step explanation:

Part a

The system of hypothesis on this case would be:

Null hypothesis: $\mu \leq \mu_0$

Alternative hypothesis: $\mu > \mu_0$

Where $\mu_0$ is the value that we want to test.

In order to find the critical value we need to find first the degrees of freedom, on this case that is given df=20. Since its an upper tailed test we need to find a value a such that:

$P(t_{20}>a) = 0.1$

And we can use excel in order to find this value with this function: "=T.INV(0.9,20)". The 0.9 is because we have 0.9 of the area on the left tail and 0.1 on the right.

The critical value on this case would be $t_{crit}=1.325$

Part b

The system of hypothesis on this case would be:

Null hypothesis: $\mu \geq \mu_0$

Alternative hypothesis: $\mu < \mu_0$

Where $\mu_0$ is the value that we want to test.

In order to find the critical value we need to find first the degrees of freedom, given by:

$df=n-1=15-1=14$

Since its an lower tailed test we need to find b value a such that:

$P(t_{14}$

And we can use excel in order to find this value with this function: "=T.INV(0.1,14)". The 0.1 is because we have 0.1 of the area accumulated on the left of the distribution.

The critical value on this case would be $t_{crit}=-1.345$

Part c

The system of hypothesis on this case would be:

Null hypothesis: $\mu = \mu_0$