Hi just to verify for estimation of sum for 148+15. Is it 100?

Mr. Romero is buying a cell phone that has a regular price of $649. The cell phone is on sale for 15% off the regular price. Wha

Dafna11 [192]

Answer:

$15 \div 100 \times 649 = 97.35 \\ 649 - 97.35 = 551.65$

therefore the new price is $551.65

6 0

2 years ago

Can someone help, me

Radda [10]

The answer to #5 is 33.

8 0

3 years ago

Read 2 more answers

A tobacco company claims that the amount of nicotene in its cigarettes is a random variable with mean 2.2 and standard deviation

Aleksandr-060686 [28]

Answer:

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 2.2, \sigma = 0.3, n = 100, s = \frac{0.3}{\sqrt{100}} = 0.03$

What is the approximate probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true?

This is 1 subtracted by the pvalue of Z when X = 3.1. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{3.1 - 2.2}{0.03}$

$Z = 30$

$Z = 30$ has a pvalue of 1.

1 - 1 = 0

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

4 0

3 years ago

A spinner has 10 equally sized sections, 5 of which are gray and 5 of which are blue. The spinner is spun twice. What is the pro

zhenek [66]

Answer:

$P(Gray\ and\ Blue) = \frac{1}{4}$

Step-by-step explanation:

Given

$Sections = 10$

$n(Gray) = 5$

$n(Blue) = 5$

Required

Determine P(Gray and Blue)

Using probability formula;

$P(Gray\ and\ Blue) = P(Gray) * P(Blue)$

Calculating P(Gray)

$P(Gray) = \frac{n(Gray)}{Sections}$

$P(Gray) = \frac{5}{10}$

$P(Gray) = \frac{1}{2}$

Calculating P(Gray)

$P(Blue) = \frac{n(Blue)}{Sections}$

$P(Blue) = \frac{5}{10}$

$P(Blue) = \frac{1}{2}$

Substitute these values on the given formula

$P(Gray\ and\ Blue) = P(Gray) * P(Blue)$

$P(Gray\ and\ Blue) = \frac{1}{2} * \frac{1}{2}$

$P(Gray\ and\ Blue) = \frac{1}{4}$

3 0

3 years ago

Herman is traveling 125 miles from his house to the beach by car. Herman plans to stop for lunch when the ratio of the distance

aliya0001 [1]

Answer:

250 miles

Step-by-step explanation:

The given distance between the house and the beach is 125 miles.

Let $x$ miles be the distance between the house and the lunch stop.

So, at the time of the lunch stop, he already traveled $x$ miles, the remaining distance is the distance between the lunch stop and the beach.

Let $y$ miles be the remaining distance, so

$y=125-x.$

Give that the ratio of the distance he has traveled, $x$ , to the distance he still has to travel, $y$ , is $2:3$ ,i.e

$x:y=2:3$

$\Rightarrow \frac x y =\frac 2 3$

$\Rightarrow \frac {x}{125-x}=\frac 2 3$

$\Rightarrow 3\times x=2\times(125-x)$

$\Rightarrow 3\times x=2\times125-2\timesx$

$\Rightarrow 3 x=250-2x$

$\Rightarrow 3x-2x=250$

$\Rightarrow x=250$

Hence, the distance traveled by Herman when he stops for lunch is 250 miles.

4 0

3 years ago