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vodka [1.7K]
3 years ago
13

What are the three types of frequency distribution?

Mathematics
1 answer:
Sliva [168]3 years ago
7 0

Answer:

Bivariate Frequency Distribution.

Cumulative Frequency Distribution.

Relative Frequency Distribution.

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If the federal budget for 2007 was 2.8 trillion dollars, how much money did the government spend per second? Please help!
diamong [38]

Solution:

we are given that

The federal budget for 2007 was 2.8 trillion dollars.

we have been asked to find

how much money did the government spend per second?

First we will find the number of seconds in one year

Number of seconds in 1 year=365*24*60*60=31536000 Seconds

Now we will dive the amount by the total number of seconds, we get

\frac{2,800,000,000,000}{31536000} =88787.41755 dollar

Hence the government is spending approximately 88787 dollars per second.

5 0
3 years ago
the cost of a school banquet is $95 plus $15 for each person attending. write an equation that gives total cost as a function of
AleksandrR [38]

Im sorry if I misunderstood the problem but heres my take.

Answer:

1155 for the cost of the people and 1250 for the total including the $95

Step-by-step explanation:

Multiply it ( 77 x 15 )

then add the answer to the $95 ( 95 + N )

= $N

8 0
3 years ago
What single percentage change is equivalent to a 7% decrease followed by a 22% decrease?
quester [9]

Answer:

<em>The single percentage change is 27.46%</em>

Step-by-step explanation:

<u>Percentage</u>

Suppose x is the initial number. It will be decreased by 7%.

(7/100)x=0.07x

x - 0.07x = 0.93x

This last number will be decreased by 22%.

0.93x*(22/100)=0.2046x

0.93x-0.2046x=0.7254x

The last number is 72.54% of the initial number. This is equivalent to a single decrease of:

100% - 72.54% = 27.46%

The single percentage change is 27.46%

5 0
3 years ago
Clark and Lana take a 30-year home mortgage of $128,000 at 7.8%, compounded monthly. They make their regular monthly payments fo
svp [43]

Answer:

Step-by-step explanation:

From the given information:

The present value of the house = 128000

interest rate compounded monthly r = 7.8% = 0.078

number of months in a year n= 12

duration of time t = 30 years

To find their regular monthly payment, we have:

PV = P \begin {bmatrix}  \dfrac{1 - (1 + \dfrac{r}{n})^{-nt}}{\dfrac{r}{n}}    \end {bmatrix}

128000 = P \begin {bmatrix}  \dfrac{1 - (1 + \dfrac{0.078}{12})^{- 12*30}}{\dfrac{0.078}{12}}    \end {bmatrix}

128000 = 138.914 P

P = 128000/138.914

P = $921.433

∴ Their regular monthly payment P = $921.433

To find the unpaid balance when they begin paying the $1400.

when they begin the payment ,

t = 30 year - 5years

t= 25 years

PV= 921.433 \begin {bmatrix}  \dfrac{1 - (1 - \dfrac{0.078}{12})^{25*30}}{\dfrac{0.078}{12}}    \end {bmatrix}

PV = $121718.2714

C) In order to estimate how many payments of $1400  it will take to pay off the loan, we have:

121718.2714 =  \begin {bmatrix}  \dfrac{1300  (1 - \dfrac{12.078}{12}))^{-nt}}{\dfrac{0.078}{12}}    \end {bmatrix}

121718.2714 = 200000  \begin {bmatrix}  (1 - \dfrac{12.078}{12}))^{-nt}   \end {bmatrix}

\dfrac{121718.2714}{200000 } =  \begin {bmatrix}  (1 - \dfrac{12.078}{12}))^{-nt}   \end {bmatrix}

0.60859 =  \begin {bmatrix}  (1 - \dfrac{12}{12.078}))^{nt}   \end {bmatrix}

0.60859 = (0.006458)^{nt}

nt = \dfrac{0.60859}{0.006458}

nt = 94.238 payments is required to pay off the loan.

How much interest will they save by paying the loan using the number of payments from part (c)?

The total amount of interest payed on $921.433 = 921.433 × 30(12) years

= 331715.88

The total amount paid using 921.433 and 1300 = (921.433 × 60 )+( 1300 + 94.238)

= 177795.38

The amount of interest saved = 331715.88  - 177795.38

The amount of interest saved = $153920.5

6 0
4 years ago
Solve the equation for x
PIT_PIT [208]

Answer:

              \bold{x=\frac{-6+z}6=-1+\frac z6}

Step-by-step explanation:

\bold{x-6+5x=-12+z}\\{}\quad +6\qquad\quad\ +6\\{\quad}\bold{6x\ =\ -6+z}\\{}\quad\div6\quad\ \ \div6\\{}\quad \bold{x=\frac{-6+z}6=-1+\frac z6}

7 0
4 years ago
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