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3 years ago

. Solve. 2/3 of 3/5 plese help

Mathematics

1 answer:

olasank [31]3 years ago

7 0

Answer:

10/9

Step-by-step explanation:

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Solve it please<br><br> 72:8

amm1812

9:1 because you divide both sides by 8. Hope this helps :)

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3 years ago

The population of a town is 13,000. It decreases at a rate of 5% per year. In about how many years will the population be less t

just olya [345]

This is an exponential decay problem.

Using the equation Y = a *(1-rate)^time

where Y is the future value given as 12,000 and a is the starting value given as 13,000.

The rate is also given as 5%.

The equation becomes:

12,000 = 13,000(1-0.05)^x

12,000 = 13,000(0.95)^x

Divide each side by 13000:

12000/13000 = 0.95^x/13000

12/13 = 0.95^x

Use the natural log function:

x = ln(12/13) / ln(0.95)

x = 1.56 years. ( this will equal 12,000

Round to 2 years it will be less than 12000.

3 0

4 years ago

PLEASE PLEASE HELP ME

andreyandreev [35.5K]

Answer:

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

If l || m, classify the marked angle pair and give their relationship, then solve for X.

pickupchik [31]

Alternate exterior angles(AEA).
Given their relationship the angles are congruent.
8x-71=5x+7
3x-71=7
3x=78
X=26

6 0

3 years ago

Recent census data indicated that 14.2% of adults between the ages of 25 and 34 live with their parents. A random sample of 125

kicyunya [14]

Answer:

The probability is $P(14 < X < 20 ) = 0.5354$

Step-by-step explanation:

From the question we are told that

The proportion that live with their parents is $\r p = 0.142$

The sample size is n = 125

Given that there are two possible outcomes and that this outcomes are independent of each other then we can say the Recent census data follows a Binomial distribution

i.e

$X \ \~ \ B( \mu , \sigma )$

Now the mean is evaluated as

$\mu = n * \r p$

$\mu = 125 * 0.142$

$\mu = 17.75$

Generally the proportion that are not staying with parents is

$\r q = 1 - \r p$

= > $\r q = 0.858$

The standard deviation is mathematically evaluated as

$\sigma = \sqrt{n * \r p * \r q }$

$\sigma = \sqrt{ 125 * 0.142 * 0.858 }$

$\sigma = 3.90$

Given the n is large then we can use normal approximation to evaluate the probability as follows

$P(14 < X < 20 ) = P( \frac{ 14 - 17.75}{3.90}$

Now applying continuity correction

$P(14 < X < 20 ) = P( \frac{ 13.5 - 17.75}{3.90} < \frac{ X - \mu }{\sigma } < \frac{ 19.5 - 17.75}{3.90} )$

Generally

$\frac{ X - \mu }{\sigma } = Z ( The \ standardized \ value \ of X )$

$P(14 < X < 20 ) = P( \frac{ 13.5 - 17.75}{3.90} < Z< \frac{ 19.5 - 17.75}{3.90} )$

$P(14 < X < 20 ) = P( -1.0897 < Z< 0.449 } )$

$P(14 < X < 20 ) = P( Z< 0.449 ) - P(Z < -1.0897)$

So for the z - table

$P( Z< 0.449 ) = 0.67328$

$P(Z < -1.0897) = 0.13792$

$P(14 < X < 20 ) = 0.67328 - 0.13792$

$P(14 < X < 20 ) = 0.5354$

6 0

3 years ago