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Alexxx [7]
3 years ago
6

Plz help help❤️Urgently

Mathematics
1 answer:
ICE Princess25 [194]3 years ago
8 0

Answer:

.15 or 3/20

Step-by-step explanation:

(3/4)/5=.15

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Melissa makes necklaces and sells them online. She charges $88 per necklace. Her monthly expenses are $3,745. How many necklaces
Dovator [93]
She will need to sell 18 necklaces a month to reach 1650
4 0
4 years ago
Read 2 more answers
Multiplying positive and negative fractions how do you do it ​
SashulF [63]

Answer:

I GOTCHU BAEE

Step-by-step explanation:

its just like regular negative numbers

but with fractions

see:

-7/12 x 1/2 = -7/24

negative negates the negative

negative negates the positive

positive keeps the positive positive

6 0
3 years ago
Suppose a $75,500 mortgage is to be amortized at 6.5% interest. Find the total amount of interest that would be paid for a 40 ye
MatroZZZ [7]
Use the formula of the present value of annuity ordinary to find the monthly payment of the loan
The formula is
Pv=pmt [(1-(1+r/k)^(-kn))÷(r/k)]
Pv present value 75500
PMT monthly payment?
R interest rate 0.065
K compounded monthly 12
N time 40 years
So we need to solve for pmt
PMT=Pv÷[(1-(1+r/k)^(-kn))÷(r/k)]
PMT=75,500÷((1−(1+0.065÷12)^(
−12×40))÷(0.065÷12))
=442.02 (this is the monthly payment)

Now find the amount of interest
Total interest=total paid-present value

Present value=75500
Total paid
442.02×12months×40years
=212,169.6

Total interest=212,169.6−75,500
=136,669.6

The answer is 136,669.6


3 0
4 years ago
Using the pattern learned for the perfect squares, what is the product?
anyanavicka [17]

Answer:

9x^2-30xy^2+25y^4

Step-by-step explanation:

3 0
3 years ago
From a random sample of 20 bars selected at random from those produced, calculations gave a mean weight of = 52.46 grams and sta
umka21 [38]

Answer:

(52.30 ; 52.62)

Step-by-step explanation:

Given :

Sample size, n = 20

Mean, xbar = 52.46

Standard deviation, s = 0.42

We assume a t - distribution

The 90% confidence interval

The confidence interval relation :

C.I = xbar ± Tcritical * s/√n

To obtain the Tcritical value :

Degree of freedom, df = n - 1 ; 20 - 1 = 19 ; α = (1 - 0.90) /2 = 0.1/2 = 0.05

Using the T-distribution table, Tcritical = 1.729

We now have :

C.I = 52.46 ± (1.729 * 0.42/√20)

C. I = 52.46 ± 0.1624

C.I = (52.30 ; 52.62)

7 0
3 years ago
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