If w = 6 feet, x = 6 feet, y = 12 feet, and z = 11 feet, what is the area of the object?

A cat had a litter of 5 kittens. If each sex is equally likely, what is the probability that 3 or less were female kittens?

ASHA 777 [7]

Answer:

0.8125

Step-by-step explanation:

In this question, we are tasked with calculating the probability that 3 or less of her kittens were female.

Since each bsex is of likely probability, the probability of a male kitten = probability of a female kitten = 0.5

Now to calculate for 3 or less female kitten we are calcualting P(f) ≤ 3

In each case, we use the Bernoulli approximation

P(f) ≤ 3 = $5C3 m^{2} f^{3} + 5C2 m^{2} f^{3} + 5C1 m^{4} f + 5C0 m^{5} f^{0}$

Where m is the probability of a male kitten and f is the probability of having a female kitten with both values = 0.5

P(f) ≤ 3 =(0.3125) + (0.3125) + (0.15625) + (0.03125) = 0.8125

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