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4 years ago

Help PLEASE!! which number is the BIGGEST negative: -3/2 or -4/5

Mathematics

2 answers:

Vilka [71]4 years ago

8 0

Answer:

-3/2 would be a farther negative number than -4/5

Step-by-step explanation:

-3/2 would be in decimal would be -1.5, while -4/5 would be -0.8

making the -3/2 a bigger negative number, technically a smaller number as its farther in the negative if thats what youre looking for

Hope this helped, brainliest?

NARA [144]4 years ago

3 0

The first step is to change our improper fraction to a mixed number. -3/2 = -1 1/2.
The key to figuring out what number is the largest when it comes to negatives is which one is closer to 0, or the farthest to the right if it were on a number line. in this case -4/5 is the largest number because it is the farthest to the right if it were in a number line.

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4 years ago

The maximum number of students in a classrom is 26 if there are 16 signed up for the art class

Aleks04 [339]

Answer:whats the question

Step-by-step explanation:question?

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3 years ago

4. Find the area of the<br> shaded region.

lozanna [386]

Diameter of the circle in the middle = 8 ft

Radius of the circle in the middle :

$=\tt \: \frac{Diameter }{2}$

$=\tt \frac{8}{2}$

$=\tt 4 \: ft$

Thus, the radius of the circle in the middle = 4 ft

Area of the entire circle :

$=\tt \pi {r}^{2}$

$= \tt3.14 \times 4 + 6 \times 4 + 6$

$= \tt3.14 \times 10 \times 10$

$=\tt 3.14 \times 100$

$\color{plum} = \tt314 \: ft$

Thus, the area of the entire circle = 314 ft

Area of the circle in the middle :

$=\tt \pi {r}^{2}$

$=\tt 3.14 \times 4 \times 4$

$= \tt3.14 \times 16$

$\color{plum} =\tt 50.24 \: ft$

Thus, the area of the circle in the middle = 50.24 ft

Area of the shaded portion :

= Total area of the figure - Area of the circle in middle

$=\tt 314 - 50.24$

$\color{plum} =\tt 263.76 \: ft$

▪︎Therefore, the area of the shaded portion = <u>263.76 ft</u>

4 0

3 years ago

The amount of time all students in a very large undergraduate statistics course take to complete an examination is distributed c

Anestetic [448]

Answer:

a) The mean is $\mu = 60$

b) The standard deviation is $\sigma = 9$

Step-by-step explanation:

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 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.

The probability a student selected at random takes at least 55.50 minutes to complete the examination equals 0.6915.

This means that when X = 55.5, Z has a pvalue of 1 - 0.6915 = 0.3085. This means that when $X = 55.5, Z = -0.5$

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

$-0.5 = \frac{55.5 - \mu}{\sigma}$

$-0.5\sigma = 55.5 - \mu$

$\mu = 55.5 + 0.5\sigma$

The probability a student selected at random takes no more than 71.52 minutes to complete the examination equals 0.8997.

This means that when X = 71.52, Z has a pvalue of 0.8997. This means that when $X = 71.52, Z = 1.28$

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

$1.28 = \frac{71.52 - \mu}{\sigma}$

$1.28\sigma = 71.52 - \mu$

$\mu = 71.52 - 1.28\sigma$

Since we also have that $\mu = 55.5 + 0.5\sigma$

$55.5 + 0.5\sigma = 71.52 - 1.28\sigma$

$1.78\sigma = 71.52 - 55.5$

$\sigma = \frac{(71.52 - 55.5)}{1.78}$

$\sigma = 9$

$\mu = 55.5 + 0.5\sigma = 55.5 + 0.5*9 = 55.5 + 4.5 = 60$

Question

The mean is $\mu = 60$

The standard deviation is $\sigma = 9$

6 0

3 years ago

Let X have the uniform distribution U(0, 2) and let the conditional distribution of Y , given that X = x, be U(0, x). Find the j

Elina [12.6K]

Answer:

$f(x,y) = \frac{1}{x} \frac{1}{2}= \frac{1}{2x} , 0\leq x \leq 2 , 0\leq y \leq x$

$E(Y|x) = \int_{x=y}^2 y \frac{1}{x} dx= y ln x \Big|_{x=y}^2 =y ln 2 -y ln y = y(1-lny) \$

Step-by-step explanation:

We have two random variables X and Y. $X \sim Unif(0,2)$ and given that X=x, Y has uniform distribution (0,x)

From the definition of the uniform distribution we have the densities for each random variable given by:

$f_X (x) =\frac{1}{2} , 0\leq x\leq 2$

$f_{Y|X} (y|x) = \frac{1}{x}, 0\leq y \leq x$

And on this case we can find the joint density with the following formula:

$f(x,y) = f_{Y|X}(y|x) f_X (x)$

And multiplying the densities we got this:

$f(x,y) = \frac{1}{x} \frac{1}{2}= \frac{1}{2x} , 0\leq x \leq 2 , 0\leq y \leq x$

Now with the joint density we can find the expected value E(Y|x) with the following formula:

$E(Y|x) = \int y f_{Y|X}(y|x)dx$

And replacing we got:

$E(Y|x) = \int_{x=y}^2 y \frac{1}{x} dx= y ln x \Big|_{x=y}^2 =y ln 2 -y ln y = y(1-lny) \$

5 0

4 years ago