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svlad2 [7]
3 years ago
8

Matthew opens a store credit card when he purchases a new sound system. The interest rate is 22.45%. He charges $4,120 to the ca

rd, and can pay $1,450 per month. What will the total cost of his purchase be?
Mathematics
2 answers:
Nookie1986 [14]3 years ago
6 0

Let

C----------------- > charges for the purchases to the credit card

<span>r------------------- > the interest rate    </span>

X------------------ > total cost of the purchase

 we know that

C=4120

r= 22.45%=0.2245

X=C*(1+r)

 substituting the values ​​in the formula

X=(4120)*(1+0.2245)= (4120)*(1.2245)=5044.94

 the answer is $5044.94


Leona [35]3 years ago
3 0

The answer is 4,273.70

Just took the quiz.. Other answer is wrong

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Will give brainliest please help ASAP
Y_Kistochka [10]
The answer should be f(x)=20(2.5)^x

How?

the graph crosses the f(x) axis at 20 but that's common in all 4 equations. However, when x=1, f(x)=50. We have to know try all the equations by substituting these values in ( both sides are equal).

f(x)=20(2.5)^x
substitute f(x)=50, x=1
50=20(2.5)^1
50=20*2.5
50=50
8 0
3 years ago
If we list all the possible factors (in order) of 12, we get 1, 2, 3, 4, 6, 12. This is the case because 1 x 12, 2 x 6, and 3 x
____ [38]

Answer:

1, 2, 3, 4, 6, 9, 12, 18, and 36, - 1, - 2

7 0
3 years ago
Daphne likes to ski at a resort that is open from December through April. According to a sign at the resort, 20, percent of the
Arturiano [62]

Answer:

Step-by-step explanation:

From the given information:

We can compute the  null hypothesis & the alternative hypothesis as:

{H_o}:\text{Distribution of snowfalls in her hometown is similar to claimed percentage }

{H_a}:\text{Distribution of snowfalls in her hometown is not similar to claimed percentage }

The degree of freedom = n - 1

The degree of freedom = 5 - 1

The degree of freedom = 4

At the level of significance of 0.05 and degree of freedom 4,

the rejection region = 9.488

However, we can compute the chi-square X² goodness of fit test as:

   

months  frequency (p)  observed O Expected E  Chi-square X^2= \dfrac{(O-E)^2}{E}

Dec          0.2                  16                   16                \dfrac{(16-16)^2}{16} =0      

Jan           0.250             11                   20                \dfrac{(11-20)^2}{20} =4.050      

Feb           0.200             16                  16                 \dfrac{(16-16)^2}{16} =0      

Mar           0.200             18                  16                 \dfrac{(18-16)^2}{16} =0.250

Apr           0.150               19                  12                 \dfrac{(19-12)^2}{12} =4.083

Total            1.000           80                 80                                  8.3833    

∴

The test statistics X² = 8.3833

Thus; we fail to reject the H_o since test statistics X² doesn't fall in the rejection region.

Therefore; there is sufficient evidence to conclude that the distribution of snowfalls in her hometown is not similar to the claimed percentage.

6 0
3 years ago
Is my answer correct
Leto [7]

mayhaps it is right and wrong is subjective

4 0
2 years ago
Please help! Answer as many as possible! ​
Mariulka [41]

Answer: 1. 10^3

One thousand

2. 10^5

One Hundred Thousand.

3.10000  4.2000  5. 60000

6. 7

70

700

7000

70000

7.12

120

1200

12000

120000

8.  9 × 1

9 × 10

9 × 100

9 × 1000

9 × 10000

9. value= 20

Step-by-step explanation:

4 0
2 years ago
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