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Mathematics

2 answers:

steposvetlana [31]3 years ago

4 0

Answer: $421.56+20g+\frac{k}{4}$

Step-by-step explanation:

You have the following information given in the problem above:

- She had $421.56 in her checking account.

- She she deposited <em>g</em> $20 bills (which you can rewrite: $20g$ ) and <em>k</em> quarters (which you can rewrite as folllowing: $\frac{k}{4}$ ).

Therefore, the amount of money in her account after the deposit is the sum of the amoun she had in her account plus the amoun of money she deposited.

Then, you can write the following expression:

$421.56+20g+\frac{k}{4}$

Mila [183]3 years ago

3 0

Answer:

421.56 + 20g + 1/4 k

Step-by-step explanation:

Joby had $421.56, and she deposited g $20 bills, and k quarter:

We don´t know how many $20 bills she deposited but we know it is expressed by the letter g, so she deposited $20g, and we add to that how many quarters of a dollar she added, in this case, we know that she deposited k quarters or 1/4 k dollars.

To write an expression to know the amount of money she has after the deposit we have to add:

What she had before the deposit = $421.56

The Dollars in $20 bills, we know it is = 20g

The number of dollars in a quarter of a dollar = 1/4 k

So we write:

421.56 + 20g + 1/4k

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According to the last census (2010), the mean number of people per household in the United States is LaTeX: \mu = 2.58 Assume a

Veseljchak [2.6K]

Answer:

P(2.50 < Xbar < 2.66) = 0.046

Step-by-step explanation:

We are given that Population Mean, $\mu$ = 2.58 and Standard deviation, $\sigma$ = 0.75

Also, a random sample (n) of 110 households is taken.

Let Xbar = sample mean household size

The z score probability distribution for sample mean is give by;

Z = $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

So, probability that the sample mean household size is between 2.50 and 2.66 people = P(2.50 < Xbar < 2.66)

P(2.50 < Xbar < 2.66) = P(Xbar < 2.66) - P(Xbar $\leq$ 2.50)

P(Xbar < 2.66) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{2.66-2.78}{\frac{0.75}{\sqrt{110} } }$ ) = P(Z < -1.68) = 1 - P(Z 1.68)

= 1 - 0.95352 = 0.04648

P(Xbar $\leq$ 2.50) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{2.50-2.78}{\frac{0.75}{\sqrt{110} } }$ ) = P(Z $\leq$ -3.92) = 1 - P(Z < 3.92)