Part (a):
We will assume that the total amount is Jan's saving account is "m"
Now, "withdrawal" means that she took money from her account, which means that the amount of money in her account decreased by the amount she withdrew.
Sum of negative integers to show her withdrawals = -25$ - 45$ - 75$
Amount of money remaining in her account = m - 145$
The amount of money Jan withdrew = 25$ + 45$ + 75$ = 145$

Part (b):
We will assume that the total amount is Lola's saving account is "n"
Now, "withdrawal" means that she took money from her account, which means that the amount of money in her account decreased by the amount she withdrew.
Sum of negative integers to show her withdrawals = -35$ - 55$ - 65$
Amount of money remaining in her account = n - 155$
The amount of money Lola withdrew = 35$ + 55$ + 65$ = 155$

7 0

2 years ago

Suppose that we wish to assess whether more than 60 percent of all U.S. households in a particular income class bought life insu

IceJOKER [234]

Answer:

A. 0.0049

B. Yes

Step-by-step explanation:

Sample proportion = 0.64

N = 1000

Population proportion = 0.60

We solve for standard deviation

= √p(1-p)/n

= √0.60(1-0.60)/1000

= √0.60x0.40/1000

= √0.00024

= 0.0155

The probability of sample >=0.64

Z>=0.64-0.60/0.0155

Z >= 0.04/0.0155

So z >= 2.5806

Using excel this equal to 0.0049

0.0049 is probability of sample proportion being 0.64 at least.

This answer in a shows that than 60% of households in the united states income class purchased life insurance last year.

3 0

2 years ago