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zaharov [31]
3 years ago
14

What date would be four months prior to January 31st?

Mathematics
1 answer:
Verdich [7]3 years ago
4 0

Answer:

sept 30

Step-by-step explanation:

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How much money would you have if you invested $3000 at 6.25% annual interest, compounded daily for 2 years?
grandymaker [24]

Answer:

A\simeq3399.41

Step-by-step explanation:

The amount formula in compound interest is:

A=P(1+\frac{r}{n} )^{nt}

where:

P = principal amount

r = annual interest

n = number of compounding periods

t = number of years

We already know that:

P = $3000

r = 6.25\% = \frac{6.25\%}{100\%}=0.0625

t = 2

n = 365

Then,

A=3000(1+\frac{0.0625}{365} )^{(365)(2)}\\\\A=3000(1+\frac{0.0625}{365} )^{730}\\\\A=3399.408982\\\\A\simeq3399.41

5 0
3 years ago
X to the 5th power and y to the 6th power over xy to the 2nd power ​
never [62]

\frac{x^5y^6}{(xy)^2}

\frac{x^5y^6}{x^2y^2}

x^{5-2}y^{6-2}

x^3y^4

6 0
4 years ago
John davis, a manager of a supermarket, wants to estimate the proportion of customers who use food stamps at his store. he has n
8_murik_8 [283]

Answer:

1509

Step-by-step explanation:

We know that, formula for number of samples is,

n= \frac{z^2 \times p \times q}{SE^2}

When nothing is given, we take p=0.5 and we know that q = 1-p=0.5.

SE = 0.03 (the true proportion should remain withing 3%)

Putting the values we get,

n= \frac{(2.33)^2 \times 0.5 \times 0.5}{(0.03)^2}

n= \frac{5.4289 \times 0.5 \times 0.5}{0.03 \times 0.03} \approx 1509


Therefore, the sample required to estimate the true proportion to within 3 percentage points is 1509.

7 0
4 years ago
HELPPPPPPPPPP ASAPPPPPP!
IrinaVladis [17]
I think it’s B because of the logical math explanation
7 0
3 years ago
Read 2 more answers
This is multiplying fractions and i have to show work
grandymaker [24]

To solve:

\frac{2}{3}\times(-4\frac{1}{3})

Let us convert the mixed fraction into fraction first,

\frac{2}{3}\times(-\frac{13}{3})

On simplification we get,

-\frac{26}{9}

Hence, the answer is,

-\frac{26}{9}

7 0
1 year ago
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