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2 years ago

127.86 +78.95 +23.09

Mathematics

2 answers:

rjkz [21]2 years ago

4 0

Answer:

229.9

Step-by-step explanation:

Reika [66]2 years ago

3 0

Answer:

229.9

Step-by-step explanation:

Uh easyyyyy lol

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Jessica bought 2 slices of pizza and a soda for $6.25. Alfredo bought 3 slices of pizza and a soda

Ira Lisetskai [31]

Answer:

A) 2.50

Step-by-step explanation:

subtract 8.75 and 6.25 to get the cost of 1 pizza slice since Alfredo bought 1 more pizza than Jessica.

4 0

2 years ago

Read 2 more answers

You deposit $975.52 in a bank account. Find the balance after 4 years if the account pays 7.25% annual interest compounded quart

sladkih [1.3K]

Answer:

The final ballance will be $1300.37.

Step-by-step explanation:

In this case we have a compounded interest, in order to calculate the final balance we need to use the following formula:

S = P(1 + r/n)^(n*t)

Where S is the final balance, P is the initial investment, r is the rate of interest, t is the time and n is the rate at which it is compounded. Since we have all the values we can directly apply to the formula as follows:

S = 975.52*(1 + 0.0725/4)^(4*4)

S = 975.52*(1.018125)^(16)

S = 975.52*1.333

S = 1300.37

The final ballance will be $1300.37.

4 0

3 years ago

The National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time requir

Paha777 [63]

Answer:

95% confidence interval for the mean time required to earn a bachelor’s degree by all college students is [5.10 years , 5.20 years].

Step-by-step explanation:

We are given that the National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time required to earn their bachelor’s degrees. The mean was 5.15 years and the standard deviation was 1.68 years respectively.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean time = 5.15 years

$\sigma$ = sample standard deviation = 1.68 years

n = sample of college graduates = 4400

$\mu$ = population mean time

Here for constructing 95% confidence interval we have used One-sample z test statistics although we are given sample standard deviation because the sample size is very large so at large sample values t distribution also follows normal.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5%

level of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95