Which quadrant would each ordered pair be in?

What formulas do I use to find surface area of cube, rectangular prism, trapezodial prism, triangular prism

-BARSIC- [3]

Cube: Area cubed
Rectangular Prism: A=(wl+hl+hw)

7 0

3 years ago

You are dealt one card from a 52 card deck. Then the card is replaced in the deck, the deck is shuffled, and you draw again. Fin

lora16 [44]

Answer:

3/52

Step-by-step explanation:

There are 13 types of cards (A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K), out of which 3 (J, Q, K) are face cards. This means that the probability for the first half is 3/13. There are 4 suites (Club, Spade, Hearts, and Diamonds). This means the probability of attaining a club is 1/4. 3/13 * 1/4 = 3/52

4 0

3 years ago

(10 points)Assume IQs of adults in a certain country are normally distributed with mean 100 and SD 15. Suppose a president, vice

vesna_86 [32]

Answer:

0.0139 = 1.39% probability that the president will have an IQ of at least 107.5 and that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

Step-by-step explanation:

To solve this question, we need to use the binomial and the normal probability distributions.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Probability the president will have an IQ of at least 107.5

IQs of adults in a certain country are normally distributed with mean 100 and SD 15, which means that $\mu = 100, \sigma = 15$

This probability is 1 subtracted by the p-value of Z when X = 107.5. So

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

$Z = \frac{107.5 - 100}{15}$

$Z = 0.5$

$Z = 0.5$ has a p-value of 0.6915.

1 - 0.6915 = 0.3085

0.3085 probability that the president will have an IQ of at least 107.5.

Probability that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

First, we find the probability of a single person having an IQ of at least 130, which is 1 subtracted by the p-value of Z when X = 130. So

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

$Z = \frac{130 - 100}{15}$

$Z = 2$

$Z = 2$ has a p-value of 0.9772.

1 - 0.9772 = 0.0228.

Now, we find the probability of at least one person, from a set of 2, having an IQ of at least 130, which is found using the binomial distribution, with p = 0.0228 and n = 2, and we want:

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{2,0}.(0.9772)^{2}.(0.0228)^{0} = 0.9549$

$P(X \geq 1) = 1 - P(X = 0) = 0.0451$

0.0451 probability that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

What is the probability that the president will have an IQ of at least 107.5 and that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130?

0.3085 probability that the president will have an IQ of at least 107.5.

0.0451 probability that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

Independent events, so we multiply the probabilities.

0.3082*0.0451 = 0.0139

0.0139 = 1.39% probability that the president will have an IQ of at least 107.5 and that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

8 0

3 years ago

Maria wraps the gift she is taking to a friend’s birthday party. She doesn’t have much paper, so she does not overlap any of the

Elina [12.6K]

Answer:

Total paper required to wrap the gift without any overlaps: $2700\ cm^{2}$

Step-by-step explanation:

Here, we need to find the total paper required without any sides overlapping to wrap the gift.

The gift is of cuboid type.

Given the following:

Length = 15 cm

Width = 30 cm

Height = 20 cm

Please refer to the attached figure.

We can infer that to find the paper required, we actually need to the find the total surface area of the cuboid.

Because the gift wrap will be done on the faces of gift (which is of cuboid shape).

Formula for Surface Area of Cuboid:

$\text {Area = } 2 \times (\text{Length}\times \text{Width} + \text{Width}\times \text{Height} + \text{Length}\times \text{Height} )\\\\\Rightarrow 2 \times (15 \times 30 + 30 \times 20 + 15 \times 20)\\\Rightarrow 2 \times (450 + 600 + 300)\\\Rightarrow 2 \times (1350)\\\Rightarrow 2700 cm^{2}$

Hence, total paper required to wrap the gift without any overlaps: $2700 cm^{2}$

8 0

3 years ago

A cylindrical can and a cone-shaped paper cup have the same radius and weight. If you use the cone to fill the van with water ho

Marta_Voda [28]

Answer:

3 cones

Step-by-step explanation:

It is given that there is a cylindrical shaped can and also a coned shaped paper cup. The radius and height of both the cylindrical can and the cone shaped paper cup are same.

So when we fill the cylindrical can with the full coned paper cup, it will take three cones of the paper cup to fill the can with water. This because the volume of a cone is $\text{one-third}$ of the volume of a cylinder that have the same base radius and height.

3 0

3 years ago