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1 month ago

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Jackie has cheering practice on Thursday and Friday if it doesn't rain. Thursday there is a 20% chance of rain and Friday there

is a 40% chance of rain. What percent chance is there that it will rain on both Thursday and Friday? IF YOU DON'T SHOW YOUR WORK I WILL REPORT YOUR ANSWER.

2 answers:

AnnZ [12.3K]1 month ago

5 0

So, Thursday has a 20% probability or 1/5 and Friday shows 40% or 2/5; multiplying them gives 1x2=2 and 5x5=25, resulting in a 2/25 chance, equivalent to an 8% likelihood.

zzz [12.3K]1 month ago

4 0

Okay, 20% converts to a fraction of 1/5 and 40% turns into 2/5.

So, 1/5 multiplied by 2/5 results in 2/25.

The fractional form of 2/25 is 8%.

Thus, you should conclude that the probability of rain on both Thursday and Friday is 8%!

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2 months ago

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Answer:

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Step-by-step explanation:

Considerando la ecuación dada $(\sqrt[3]{128} )^{x}\\$

De acuerdo con una de las leyes de índices,

$(\sqrt[a]{m} )^{b}\\= (\sqrt{m})^\frac{b}{a}$

Aplicando esta ley a la pregunta;

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3 months ago

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Inessa [12570]

Answer:

a) $P(X>20)=P(\frac{X-\mu}{\sigma}>\frac{20-\mu}{\sigma})=P(Z>\frac{20-15}{3.5})=P(z>1.43)$

The probability can be determined using the complement rule, with the standard normal distribution, an excel sheet, or a calculator.

$P(z>1.43)=1-P(z$

b) $P(X$

This probability can also be calculated using the normal standard distribution, an excel sheet, or a calculator.

$P(z$

c) $P($

For this one, the probability can likewise be derived from the standard normal distribution, excel, or a calculator, with specific adjustments:

$P(-1.43$

Step-by-step explanation:

Previous concepts

Normal distribution refers to a symmetric probability distribution centered around the mean, indicating that occurrences near the mean are more common than those far from it.

The Z-score is a statistic that represents a value's relationship to the average of a set of values, expressed in terms of how many standard deviations it is away from the mean.

Part a

Let X denote the random variable representing the lengths within a population, and for our case, the distribution for X is as follows:

$X \sim N(15,3.5)$

Where $\mu=15$ and $\sigma=3.5$

We seek the probability:

$P(X>20)$

The most effective way to solve this is by leveraging the normal distribution and the corresponding Z-score:

$z=\frac{x-\mu}{\sigma}$

By applying this formula, we can find the probability:

$P(X>20)=P(\frac{X-\mu}{\sigma}>\frac{20-\mu}{\sigma})=P(Z>\frac{20-15}{3.5})=P(z>1.43)$

Again, this probability can be obtained either using the complement rule, the standard normal distribution, or a calculator.

$P(z>1.43)=1-P(z$

Part b

$P(X$

This probability can also be computed using either the normal standard distribution, an excel sheet, or a calculator.

$P(z$

Part c

$P($

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$P(-1.43$

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Leona [12618]

Response:

a

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b

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c

At 35 minutes since the coffee was placed on the counter, the temperature is 68, and it will drop by approximately 0.75 in the next 30 seconds.

Detailed explanation:

According to the question, we are informed that

The temperature of a cup of coffee on a counter is represented by the equation

$H= f(t)$

where t indicates the time in minutes since the coffee was set on the counter.

In general, f(t)' denotes the derivative of f(t), indicating the change in temperature over time for the coffee

As the coffee rests on the counter, its temperature tends to decrease after a few minutes; thus, f(t)' is negative.

The temperature is measured in degrees Celsius while time is expressed in minutes.

Therefore, the change in temperature with respect to time, f(35)', will be given in degrees Celsius per minute (i.e., degC/min).

The question states that

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which means the rate of temperature change after 35 minutes is 1.5.

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$R = |f(35)'| * t$

=> $R = 1.5 * 0.5$

=> $R = 0.75$

Thus

At 35 minutes since the coffee was placed on the counter, its temperature is 68 and will drop by roughly 0.75 in the next 30 seconds.

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