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How Do Solve 3/4=3/8x-3/2 And Show All Your Work

lubasha [3.4K]

It's too hard for me to show all the work, but you can find a common denominator in 8.

Change 3/4 to 6/8 and -3/2 to -12/8.

Now that all the fractions have the same denominator, you can multiply the whole problem by 8 and cancel out all denominators.

You are left with 6=3x-12

Add 12 to 6 and cancel out on the right side.

18=3x

Divide by 18 by 3

x=6

3 0

3 years ago

Read 2 more answers

What is the surface area? 1 mm 7 mm 3 mm square millimeters

AveGali [126]

Answer:

0.62 cm

Step-by-step explanation:

8 0

3 years ago

A scientist begins with 250 grams of a radioactive substance. After 250 minutes, the sample has decayed to 32 grams. Rounding to

s2008m [1.1K]

Answer:

Part 1 : $A=250 e^{-0.008223t}$

Part 2 : Half life is 84 minutes ( approx )

Step-by-step explanation:

Part 1 : Suppose the function that shows the amount( in grams ) of the substance after t minutes,

$A=A_0 e^{kt}$

If t = 0 min, A = 250 grams,

$250=A_0 e^{0}$

$\implies A_0 = 250$

If t = 250, A = 32 grams,

$32 = A_0 e^{250k}$

$32 = 250 e^{250k}$

$0.128 = e^{250k}$

Taking ln both sides,

$\ln(0.128) = 250k$

$\implies k =\frac{\ln(0.128)}{250}=-0.008223$

Hence, the equation that shows this situation,

$A=250 e^{-0.008223t}$

Part 2 : If A = 250/2 = 125,

$125 = 250 e^{-0.008223t}$

$0.5 = e^{-0.008223t}$

Taking ln both sides,

$\ln(0.5) = -0.008223t$

$\implies t =\frac{\ln(0.5)}{-0.008223}\approx 84$

Therefore, the half life of the substance would be 84 minutes.

8 0

3 years ago

Suppose a geyser has a mean time between eruptions of 72 minutes. Let the interval of time between the eruptions be normally dis

nikitadnepr [17]

Answer:

(a) The probability that a randomly selected time interval between eruptions is longer than 82 minutes is 0.3336.

(b) The probability that a random sample of 13-time intervals between eruptions has a mean longer than 82 minutes is 0.0582.

(c) The probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 minutes is 0.0055.

(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.

(e) The population mean must be more than 72, since the probability is so low.

Step-by-step explanation:

We are given that a geyser has a mean time between eruptions of 72 minutes.

Also, the interval of time between the eruptions be normally distributed with a standard deviation of 23 minutes.

(a) Let X = the interval of time between the eruptions

So, X ~ N( $\mu=72, \sigma^{2} =23^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean time = 72 minutes

$\sigma$ = standard deviation = 23 minutes

Now, the probability that a randomly selected time interval between eruptions is longer than 82 minutes is given by = P(X > 82 min)

P(X > 82 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{82-72}{23}$ ) = P(Z > 0.43) = 1 - P(Z $\leq$ 0.43)

= 1 - 0.6664 = 0.3336

The above probability is calculated by looking at the value of x = 0.43 in the z table which has an area of 0.6664.

(b) Let $\bar X$ = sample mean time between the eruptions

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean time = 72 minutes

$\sigma$ = standard deviation = 23 minutes

n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between eruptions has a mean longer than 82 minutes is given by = P( $\bar X$ > 82 min)

P( $\bar X$ > 82 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{82-72}{\frac{23}{\sqrt{13} } }$ ) = P(Z > 1.57) = 1 - P(Z $\leq$ 1.57)

= 1 - 0.9418 = 0.0582

The above probability is calculated by looking at the value of x = 1.57 in the z table which has an area of 0.9418.

(c) Let $\bar X$ = sample mean time between the eruptions

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean time = 72 minutes

$\sigma$ = standard deviation = 23 minutes

n = sample of time intervals = 34

Now, the probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 minutes is given by = P( $\bar X$ > 82 min)

P( $\bar X$ > 82 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{82-72}{\frac{23}{\sqrt{34} } }$ ) = P(Z > 2.54) = 1 - P(Z $\leq$ 2.54)

= 1 - 0.9945 = 0.0055

The above probability is calculated by looking at the value of x = 2.54 in the z table which has an area of 0.9945.

(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.

(e) If a random sample of 34-time intervals between eruptions has a mean longer than 82 minutes, then we conclude that the population mean must be more than 72, since the probability is so low.

6 0

3 years ago

How do I solve these trigonometric functions?

aleksandr82 [10.1K]

Answer:

see attached

Step-by-step explanation:

6 0

3 years ago