Cassie's cat weight's 3 1/2 pounds. her dog weighs 4 times as much as her cat. how much does her dog weigh?

Levart [38]

So $39.50 is the price when you are paying full price, which can be represented by 100%. If you’re paying 25% less, than you are only paying 75% of the original price. Simply multiply $39.50 by 75%(expresses as 0.75) to determine the sale price. 39.50(.75)=29.62 The closest value to that is $29.50, so D is the answer.

6 0

3 years ago

The probability that a biased dice will land on a six is 0.4. Marie is going to throw the dice 400 times. Work out an estimate f

steposvetlana [31]

I’m pretty sure it’s 16...?

4 0

2 years ago

The BlueSky Airlines has sold tickets for his plane to travel from Savannah to Atlanta. This plane carries 115 passengers. Howev

oee [108]

Answer:

it would be day 4 because if u add all of the tickets up once you get to day 5 it goes over 115

Step-by-step explanation:

8 0

3 years ago

Rodger has 47 cars . Can he group the cars in more than two ways?

deff fn [24]

First of all, Rodger must be making some serious cash to have 47 cars. Going back to the question, he can not group the cars in more than 2 ways. He can only group it in 1 way, 47 groups of 1. Pls brainliest

4 0

2 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Anettt [7]

Answer:

Choice C: approximately 121 green beans will be 13 centimeters or shorter.

Step-by-step explanation:

What's the probability that a green bean from this sale is shorter than 13 centimeters?

Let the length of a green bean be $X$ centimeters.

$X$ follows a normal distribution with

mean $\mu = 11.2$ and
standard deviation $\sigma = 2.1$ .

In other words,

$X\sim \text{N}(11.2, 2.1^{2})$ ,

and the probability in question is $X \le 13$ .

Z-score table approach:

Find the z-score of this measurement:

$\displaystyle z= \frac{x-\mu}{\sigma} = \frac{13-11.2}{2.1} = 0.857143$ . Closest to 0.86.

Look up the z-score in a table. Keep in mind that entries on a typical z-score table gives the probability of the left tail, which is the chance that $Z$ will be less than or equal to the z-score in question. (In case the question is asking for the probability that $Z$ is greater than the z-score, subtract the value from table from 1.)

$P(X\le 13) = P(Z \le 0.857143) \approx 0.8051$ .

"Technology" Approach

Depending on the manufacturer, the steps generally include:

Locate the cumulative probability function (cdf) for normal distributions.
Enter the lower and upper bound. The lower bound shall be a very negative number such as $-10^{9}$ . For the upper bound, enter $13$
Enter the mean and standard deviation (or variance if required).
Evaluate.

For example, on a Texas Instruments TI-84, evaluating $\text{normalcdf})(-1\text{E}99,\;13,\;11.2,\;2.1 )$ gives $0.804317$ .

As a result,

$P(X\le 13) = 0.804317$ .

Number of green beans that are shorter than 13 centimeters:

Assume that the length of green beans for sale are independent of each other. The probability that each green bean is shorter than 13 centimeters is constant. As a result, the number of green beans out of 150 that are shorter than 13 centimeters follow a binomial distribution.

Number of trials $n$ : 150.
Probability of success $p$ : 0.804317.

Let $Y$ be the number of green beans out of this 150 that are shorter than 13 centimeters. $Y\sim\text{B}(150,0.804317)$ .

The expected value of a binomial random variable is the product of the number of trials and the probability of success on each trial. In other words,

$E(Y) = n\cdot p = 150 \times 0.804317 = 120.648\approx 121$

The expected number of green beans out of this 150 that are shorter than 13 centimeters will thus be approximately 121.

7 0

3 years ago