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4 years ago

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A recipe calls for 3/1/3 cups of flour. Charlie only has 1/3 -cup measuring cup. How many times will Charlie need to fill the 1/

3 -cup measuring cup? Set up the initial problem on the left side of the equation and the solution on the right side of the equation..

1 answer:

Lena [83]4 years ago

8 0

10... Charlie will need 10 1/3 cups of flour. 1/3x9=3 so 9+1=10

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Evaluate (a + b)2 for a = 2 and b = 3.<br> 1.) 10<br> 2.)13<br> 3.)25

Fittoniya [83]

(a + b)2 Plug in the a and b values
(2 + 3)2 Add 2 and 3
(5)2 Multiply 5 and 2
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When a = 2 and b = 3, (a + b)2 will equal 1.) 10.

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3 years ago

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Tacoma's population in 2000 was about 200 thousand, and had been growing by about 9% each year. a. Write a recursive formula for

KIM [24]

Answer:

a) The recurrence formula is $P_n = \frac{109}{100}P_{n-1}$ .

b) The general formula for the population of Tacoma is

$P_n = \left(\frac{109}{100}\right)^nP_{0}$ .

c) In 2016 the approximate population of Tacoma will be 794062 people.

d) The population of Tacoma should exceed the 400000 people by the year 2009.

Step-by-step explanation:

a) We have the population in the year 2000, which is 200 000 people. Let us write $P_0 = 200 000$ . For the population in 2001 we will use $P_1$ , for the population in 2002 we will use $P_2$ , and so on.

In the following year, 2001, the population grow 9% with respect to the previous year. This means that $P_0$ is equal to $P_1$ plus 9% of the population of 2000. Notice that this can be written as

$P_1 = P_0 + (9/100)*P_0 = \left(1-\frac{9}{100}\right)P_0 = \frac{109}{100}P_0$ .

In 2002, we will have the population of 2001, $P_1$ , plus the 9% of $P_1$ . This is

$P_2 = P_1 + (9/100)*P_1 = \left(1-\frac{9}{100}\right)P_1 = \frac{109}{100}P_1$ .

So, it is not difficult to notice that the general recurrence is

$P_n = \frac{109}{100}P_{n-1}$ .

b) In the previous formula we only need to substitute the expression for $P_{n-1}$ :

$P_{n-1} = \frac{109}{100}P_{n-2}$ .

Then,

$P_n = \left(\frac{109}{100}\right)^2P_{n-2}$ .

Repeating the procedure for $P_{n-3}$ we get

$P_n = \left(\frac{109}{100}\right)^3P_{n-3}$ .

But we can do the same operation n times, so

$P_n = \left(\frac{109}{100}\right)^nP_{0}$ .

c) Recall the notation we have used:

$P_{0}$ for 2000, $P_{1}$ for 2001, $P_{2}$ for 2002, and so on. Then, 2016 is $P_{16}$ . So, in order to obtain the approximate population of Tacoma in 2016 is

$P_{16} = \left(\frac{109}{100}\right)^{16}P_{0} = (1.09)^{16}P_0 = 3.97\cdot 200000 \approx 794062$

d) In this case we want to know when $P_n>400000$ , which is equivalent to

$(1.09)^{n}P_0>400000$ .

Substituting the value of $P_0$ , we get

$(1.09)^{n}200000>400000$ .

Simplifying the expression:

$(1.09)^{n}>2$ .

So, we need to find the value of $n$ such that the above inequality holds.

The easiest way to do this is take logarithm in both hands. Then,

$n\ln(1.09)>\ln 2$ .

So, $n>\frac{\ln 2}{\ln(1.09)} = 8.04323172693$ .

So, the population of Tacoma should exceed the 400 000 by the year 2009.

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3 years ago

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kari74 [83]

The answer is C. 16/55 because P(Q and R) = P (Q) . P (R) (for independent events Q and R)

So, the prob. is 4/5 * 4/11 = 16/55 = D.

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3 years ago

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Yesterday, Kevin had m baseball cards. Today, he gave 9 away. Using m, write an expression for the number of cards Kevin left

Scorpion4ik [409]

Answer:

m-9

Step-by-step explanation:

m represents how much cards he used to have he gave away 9 so u subtract

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3 years ago

$14.65 plus what equals $65.00?

ziro4ka [17]

79.65

I used a calculator

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