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3 years ago

How to find area of a shaded region

Mathematics

2 answers:

vazorg [7]3 years ago

7 0

You’d have to know the shape and length of the sides

dezoksy [38]3 years ago

5 0

It depends on what shape it is...

You might be interested in

Can someone help me with number 50?

timama [110]

Let's take a peek at the denominators first off, and do a "prime factoring" on them.

25 = 5 * 5

27 = 3 * 3 * 3

45 = 3 * 3 * 5

the numbers in bold are repeated there more than once among the factors, and therefore we'll use them once, so our LCD will be 5 * 5 * 3 * 3 * 3, or 675.

$\bf \stackrel{Alaska}{\cfrac{4}{25}}~+~\stackrel{Texas}{\cfrac{2}{27}}~+~\stackrel{California}{\cfrac{2}{45}}~~ \stackrel{LCD~675}{\implies }~~ \cfrac{(27)4~+~(25)2~+~(15)2}{675}
\\\\\\
\cfrac{108+50+30}{675}\implies \cfrac{188}{675}$

3 0

3 years ago

Can someone help please ill give Brainlist

Maru [420]

Answer:

-1/3

Step-by-step explanation:

The common difference is how much it goes up by with each new number, so you can see that each time it goes down by 1/3. This means that the common difference is -1/3.

7 0

2 years ago

colin climbs 16 feet down into tunnel and lands on the tunnel floor. then he jumps to a platform that is 2 feet above the tunnel

Ksivusya [100]

Answer:

-14

Step-by-step explanation:

-16 + 2=-14

7 0

2 years ago

The number of months a group of middle-aged men have been on a certain heart medication is normal

Liono4ka [1.6K]

Answer:

34 %

Step-by-step explanation:

One standard deviation below ( 2 months) to the mean

covers approx 34.1 %

5 0

2 years ago

In a recent year, the scores for the reading portion of a test were normally distributed, with a mean of 22.4 and a standard dev

Dovator [93]

Using the normal distribution, we have that:

a) The probability of a student scoring less than 19 is 0.2709 = 27.09%.

b) The probability of a student scoring between 17.7 and 27.1 is 0.6424 = 64.24%.

c) The probability of a student scoring more than 33.1 is 0.0179 = 1.79%.

d) The correct option is: B. The event in part (c) is unusual because its probability is less than 0.05.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation are given, respectively, by:

$\mu = 22.4, \sigma = 5.1$

Item a:

The probability is the <u>p-value of Z when X = 19</u>, hence:

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

$Z = \frac{19 - 22.4}{5.1}$

Z = -0.67.

Z = -0.67 has a p-value of 0.2709.

The probability of a student scoring less than 19 is 0.2709 = 27.09%.

Item b:

The probability is the <u>p-value of Z when X = 27.1 subtracted by the p-value of Z when X = 17.7</u>, hence:

X = 27.1:

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

$Z = \frac{27.1 - 22.4}{5.1}$

Z = 0.92.

Z = 0.92 has a p-value of 0.8212.

X = 17.7:

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

$Z = \frac{17.7 - 22.4}{5.1}$

Z = -0.92.

Z = -0.92 has a p-value of 0.1788.

0.8212 - 0.1788 = 0.6424.

The probability of a student scoring between 17.7 and 27.1 is 0.6424 = 64.24%.

Item c:

The probability is <u>one subtracted by the p-value of Z when X = 33.1</u>, hence:

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

$Z = \frac{33.1 - 22.4}{5.1}$

Z = 2.1.

Z = 2.1 has a p-value of 0.9821.

1 - 0.9821 = 0.0179.

The probability of a student scoring more than 33.1 is 0.0179 = 1.79%.

Item d:

Probabilities less than 0.05 are unusual, hence the correct option is:

B. The event in part (c) is unusual because its probability is less than 0.05.

More can be learned about the normal distribution at brainly.com/question/28135235

#SPJ1

8 0

2 years ago