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3 years ago

Oranges cost 25 cents for 1/2 kilograms. How much would 8 kilograms of oranges cost?

2 answers:

6 0

$\dfrac{1}{2} \text{ kg} \longrightarrow 25 \text{ cents}$

Find 1 kg:
$1 \text{ kg} \longrightarrow 25 \times 2 \text{ cents}$

$1 \text{ kg} \longrightarrow 50 \text{ cents}$

Find 8 kg:
$8\text{ kg} \longrightarrow 50 \times 8 \text{ cents}$

$8\text{ kg} \longrightarrow 400 \text{ cents or } \$4.00$

Answer: 8 kg of oranges cost $4.00

lakkis [162]3 years ago

5 0

The cost of 8 kilograms of oranges cost 4$

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63.7999... (repeating) fps

Step-by-step explanation:

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2 years ago

You are working in a primary care office. Flu season is starting. For the sake of public health, it is critical to diagnose peop

Aleksandr-060686 [28]

Answer:

E. 0.11

Step-by-step explanation:

We have these following probabilities:

A 10% probability that a person has the flu.

A 90% probability that a person does not have the flu, just a cold.

If a person has the flu, a 99% probability of having a runny nose.

If a person just has a cold, a 90% probability of having a runny nose.

This can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

$P = \frac{P(B).P(A/B)}{P(A)}$

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

In this problem, we have that:

What is the probability that a person has the flu, given that she has a runny nose?

P(B) is the probability that a person has the flu. So P(B) = 0.1.

P(A/B) is the probability that a person has a runny nose, given that she has the flu. So P(A/B) = 0.99.

P(A) is the probability that a person has a runny nose. It is 0.99 of 0.1 and 0.90 of 0.90. So

$P(A) = 0.99*0.1 + 0.9*0.9 = 0.909$

What is the probability that this person has the flu?

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.1*0.99}{0.909} = 0.1089 = 0.11$

The correct answer is:

E. 0.11

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3 years ago

You have measured the systolic blood pressure of a random sample of 25 employees of a company located near you. A 95% confidence

Alla [95]

Answer:

$122 \leq \mu \leq 138$

The 95% confidence interval would be given by (122;138)

And the correct interpretation for this case is:

c. If the procedure were repeated many times, 95% of the resulting confidence intervals would contain the population mean systolic blood pressure.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % where by a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=25-1=24$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,24)".And we see that $t_{\alpha/2}=2.06$