Monica needs to increase her savings by 20%. If she has $85 already in her savings account, how much does she need to add?

Find the average rate of change over the given interval f(x)=3^x,[1,4]

nexus9112 [7]

Answer:

26

Step-by-step explanation:

Thus, f(b)−f(a)b−a=3(4)−(3(1))4−(1)=26.

7 0

3 years ago

Why are 1 4 9 16 and 25 called perfect squares

olga nikolaevna [1]

Perfect squares can be divided and multiplied by the same number to get the same number. For example 25/5=5. 5*5=25.
4/2=2, 2*2=4
1/1=1, 1*1=1
9/3=3, 3*3=9
16/4=4, 4*4=16.
Hence the perfect squares.

4 0

3 years ago

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caroline has $222 in her piggy bank. she adds $10 a week to her savings. javier has $148 in his piggy bank, and adds $12 a week

Afina-wow [57]

Answer:

392

Step-by-step explanation:

6 0

3 years ago

If the first step in the equation 2x-8=5x+3 is "subtract 2x," then in the form of a paragraph explain in complete sentences the

zubka84 [21]

2x - 8 = 5x +3
First you will subtract the 2x and you will end up with -8 = 3x + 3. Then you will need to move the 3 over by subtraction. You will end up with -11 = 3x. Now you will divide each side by 3 to get the x by itself. You end up with -11/3 = x as your final answer.

6 0

3 years ago

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(a) A lamp has two bulbs of a type with an average lifetime of 1600 hours. Assuming that we can model the probability of failure

lara [203]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The probability is $P_T= 0.4560$

b

The probability is $P_F= 0.0013$

Step-by-step explanation:

From the question we are told that

The mean for the exponential density function of bulbs failure is $\mu = 1600 \ hours$

Generally the cumulative distribution for exponential distribution is mathematically represented as

$1 - e^{- \lambda x}$

The objective is to obtain the p=probability of the bulbs failure within 1800 hours

So for the first bulb the probability will be

$P_1(x < 1800)$

And for the second bulb the probability will be

$P_2 (x< 1800)$

So from our probability that we are to determine the area to the left of 1800 on the distribution curve

Now the rate parameter $\lambda$ is mathematically represented as

$\lambda$ $= \frac{1}{\mu}$

$\lambda = \frac{1}{1600}$

The probability of the first bulb failing with 1800 hours is mathematically evaluated as

$P_1(x < 1800) = 1 - e^{\frac{1}{1600} * 1800 }$

$= 0.6753$

Now the probability of both bulbs failing would be

$P_T=P_1(x < 1800) * P_2(x < 1800)$

$= 0.6375 * 06375$

$P_T= 0.4560$

Let assume that one bulb failed at time $T_a$ and the second bulb failed at time $T_b$ then

$T_a + T_b = 1800\ hours$

The mathematical expression to obtain the probability that the first bulb failed within between zero and $T_a$ and the second bulb failed between $T_a \ and \ 1800$ is represented as

$P_F=\int_{0}^{1800}\int_{0}^{1800-x} \f{\lambda }^{2}e^{-\lambda x}* e^{-\lambda y}dx dy$

$=\int_{0}^{1800} {\lambda }e^{-\lambda x}\int_{0}^{1800-x} {\lambda } e^{-\lambda y}dx dy$

$=\int_{0}^{1800} {\frac{1}{1600} }e^{-\lambda x}\int_{0}^{1800-x} \frac{1}{1600 } e^{-\lambda y}dx dy$

$=\int_{0}^{1800} {\frac{1}{1600} }e^{-\lambda x}[e^{- \lambda y}]\left {1800-x} \atop {0}} \right. dx$

$=\int_{0}^{1800} {\frac{1}{1600} }e^{-\frac{x}{1600} }[e^{- \frac{1800 -x}{1600} }-1] dx$

$=[ {\frac{1}{1600} }e^{-\frac{1800}{1600} }-\frac{1}{1600}[e^{- \frac{x}{1600} }] \left {1800} \atop {0}} \right.$

$=[ {\frac{1}{1600} }e^{-\frac{1800}{1600} }-\frac{1}{1600}[e^{- \frac{1800}{1600} }] -[[ {\frac{1}{1600} }e^{-\frac{1800}{1600} }-\frac{1}{1600}[e^{-0}]$

$=[\frac{1}{1600} e^{-\frac{1800}{1600} } - \frac{1}{1600} e^{-0} ]$

$=0.001925 -0.000625$

$P_F= 0.0013$

4 0

3 years ago

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