- 4y - 4+ (-3) Combine like terms

How to find the other endpoint with the given endpoint and midpoint?

prohojiy [21]

To find the other endpoint with a given endpoint and midpoint, you subtract the midpoint values from the endpoint, then with whatever the answer is, you subtract that from the midpoint.

6 0

3 years ago

A city hosted a music festival that included three concerts. According to the sales database, 28% of the audience attended the f

Vitek1552 [10]

Answer:

$P(X\cap Y\cap Z)=0.05$

Step-by-step explanation:

From the question we are told that:

Percentage of audience in first concert $P(X)=0.28$

Percentage of audience in second concert $P(Y)=0.42$

Percentage of audience in third concert $P(Z)=0.30$

Audience Percentage at at-least one concert $P(X \cup Y \cup Z)=0.80$

Percentage of audience at first & second concert $P(X \cap Y)=0.10$

Percentage of audience in first & third concert $P(X \cap Z)=0.08$

Percentage of audience in second & third concert $P(Y\cap Z)=0.07$

Generally the equation for probability of attending all concerts $P(X\cap Y\cap Z)$ is mathematically given by

$P(X \cup Y \cup Z)=P(X)+P(Y)+P(Z)-P(X \cap Y)-P(X \cap Z)-P(Y\cap Z)+P(X\cap Y\cap Z)$

$P(X\cap Y\cap Z)=P(X \cup Y \cup Z)-P(X)-P(Y)-P(Z)+P(X \cap Y)+P(X \cap Z)+P(Y\cap Z)$

$P(X\cap Y\cap Z)=0.80-0.28-0.42-0.30+0.10+0.80+0.70$

$P(X\cap Y\cap Z)=0.05$

Therefore the probability that a randomly selected audient attended all the concerts

$P(X\cap Y\cap Z)=0.05$

8 0

3 years ago

Suppose a horticulturist measures the aboveground height growth rate of four different ornamental shrub species grown in a green

torisob [31]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The decision is to reject the null hypothesis at a significant level of significance $\alpha = 0.05$ There is sufficient evidence to conclude that at least one of the population mean is different from at least of the population

Step-by-step explanation:

From the question we are told that the claim is

The mean growth rates of all four species are equal.

The null hypothesis is

$H_o : \mu _1 = \mu_2 = \mu_3 = \mu_4$

Th alternative hypothesis is

$H_a: at \ least \ one \ of \ the \ means \ is \not\ equal$

From question the p-value is $p-value = 0.015$

And since the $p-value < \alpha$ so the null hypothesis will be rejected

So

The decision is to reject the null hypothesis at a significant level of significance $\alpha = 0.05$ There is sufficient evidence to conclude that at least one of the population mean is different from at least of the population