I WILL GIVE BRAINLEST!!!!!!!

Ede4ka [16]

Given the expression
-2 + 6.45z - 6 + (-3.25z)

First, classify like terms, constant with constant, coefficient-variable with coefficient-variable
-2 + 6.45z - 6 + (-3.25z)
= 6.45z + (-3.25z) - 2 - 6

Second, combining like terms
= 6.45z + (-3.25z) - 2 - 6
= 6.45z - 3.25z - 8
= 3.20z - 8

The simplest form is 3.20 - 8

8 0

3 years ago

Based on the sample of 500 people, 42% owned cats. Calculate the test statistic. Round to two decimal places.

Tanzania [10]

Answer:

z= 3.63

z for significance level = 0.05 is ± 1.645

Step-by-step explanation:

Here p = 42% = 0.42

n= 500

We formulate our null and alternative hypotheses as

H0: p= 0.42 against Ha : p> 0.42 One tailed test

From this we can find q which is equal to 1-p= 1-0.42 = 0.58

Taking p`= 0.5

Now using the z test

z= p`- p/ √p(1-p)/n

Putting the values

z= 0.5- 0.42/ √0.42*0.58/500

z= 0.5- 0.42/ 0.0220

z= 3.63

For one tailed test the value of z for significance level = 0.05 is ± 1.645

Since the calculated value does not fall in the critical region we reject our null hypothesis and accept the alternative hypothesis that more than 42% people owned cats.

4 0

3 years ago

Helppppppppppppppppppppppp

DiKsa [7]

Answer:

gotta be answer A

Step-by-step explanation:

you can search up f(x) = square root of x

7 0

3 years ago

to get your carpet cleaned in your house, Easy-Clean charges $50 cleaning fee plus $12 per room. write an equation

8_murik_8 [283]

C represents the cost. R represents the number of rooms.
C=50 + 12R

4 0

3 years ago

Read 2 more answers

An automobile manufacturer finds that 1 in every 2500 automobiles produced has a particular manufacturing defect. (a) Use a bin

Advocard [28]

Answer:

a) 0.1558 = 15.58% probability of finding 4 cars with the defect in a random sample of 7000 cars.

b) 0.1557 = 15.57% probability of finding 4 cars with the defect in a random sample of 7000 cars. These probabilities are very close, which means that the approximation works.

Step-by-step explanation:

Binomial distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

To use the Poisson approximation for the binomial, we have that:

$\mu = np$

1 in every 2500 automobiles produced has a particular manufacturing defect.

This means that $p = \frac{1}{2500} = 0.0004$

a) Use a binomial distribution to find the probability of finding 4 cars with the defect in a random sample of 7000 cars.

This is $P(X = 4)$ when $n = 7000$ . So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 4) = C_{7000,4}.(0.0004)^{4}.(0.9996)^{6996} = 0.1558$

0.1558 = 15.58% probability of finding 4 cars with the defect in a random sample of 7000 cars.

(b) The Poisson distribution can be used to approximate the binomial distribution for large values of n and small values of p.

Using the approximation:

$\mu = np = 7000*0.0004 = 2.8$ . So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 4) = \frac{e^{-2.8}*(2.8)^{4}}{(4)!} = 0.1557$

0.1557 = 15.57% probability of finding 4 cars with the defect in a random sample of 7000 cars. These probabilities are very close, which means that the approximation works.

6 0

3 years ago