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nikitadnepr [17]

3 years ago

5

20 CHILDREN WENT TO A CAFE FOR LUNCH.14 CHILDREN ATE FISH BURGERS. 10 CHILDREN ATE CHICKEN BURGERS. HOW MANY CHILDREN ATE BOTH F

ISH AND CHICKEN BURGERS?

1 answer:

Ber [7]3 years ago

6 0

If there are 20 children that went to eat lunch, and 14 of them ate fish burgers, and 10 ate chicken burgers, you could make a line graph with 20 points and color the kids who ate fish burgers yellow and the one who ate chicken burgers blue. The green where they overlapped, and 10 of them over lapped, meaning that <u>10 children ate both fish and chicken burgers. </u>Hope I helped!

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Which fractions are less than 1/2? A. 1/3 B. 4/6 C. 2/5 D. 5/8 or E. 7/12

butalik [34]

The fractions less than 1/2 are:
A. 1/3
C. 2/5

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3 years ago

If Tanisha has $1000 to invest at 7% per annum compounded semiannually, how long will it be before she has $1600? If the co

Sphinxa [80]

Answer:

Using continuous interest 6.83 years before she has $1600.

Using continuous compounding, 6.71 years.

Step-by-step explanation:

Compound interest:

The compound interest formula is given by:

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

Where A(t) is the amount of money after t years, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per unit year and t is the time in years for which the money is invested or borrowed.

Continuous compounding:

The amount of money earned after t years in continuous interest is given by:

$P(t) = P(0)e^{rt}$

In which P(0) is the initial investment and r is the interest rate, as a decimal.

If Tanisha has $1000 to invest at 7% per annum compounded semiannually, how long will it be before she has $1600?

We have to find t for which $A(t) = 1600$ when $P = 1000, r = 0.07, n = 2$

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

$1600 = 1000(1 + \frac{0.07}{2})^{2t}$

$(1.035)^{2t} = \frac{1600}{1000}$

$(1.035)^{2t} = 1.6$

$\log{1.035)^{2t}} = \log{1.6}$

$2t\log{1.035} = \log{1.6}$

$t = \frac{\log{1.6}}{2\log{1.035}}$

$t = 6.83$

Using continuous interest 6.83 years before she has $1600

If the compounding is continuous, how long will it be?

We have that $P(0) = 1000, r = 0.07$

Then

$P(t) = P(0)e^{rt}$

$1600 = 1000e^{0.07t}$

$e^{0.07t} = 1.6$

$\ln{e^{0.07t}} = \ln{1.6}$

$0.07t = \ln{1.6}$

$t = \frac{\ln{1.6}}{0.07}$

$t = 6.71$

Using continuous compounding, 6.71 years.

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