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2 years ago

1.What value goes in the blank to make this a proportion:

Mathematics

1 answer:

iragen [17]2 years ago

4 0

1. 24
2. 56
3. 15
4. 60
Hopefully this helps!

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If P(x,y) is the point on the unit circle defined by real number e, then csce =

olasank [31]

Answer:

$csc(e) = \frac{1}{y}$

Step-by-step explanation:

It is given that the point P(x, y) lies on the unit circle and is defined by real number e. We have to find csc(e) i.e cosecant of (e)

This scenario is represented in the figure attached below.

The formula of csc is:

$csc(\theta) = \frac{\text{Hypotenuse}}{\text{Opposite}}$

A right angled triangle is constructed, as shown in figure. The hypotenuse of the triangle forms the radius of the circle. Since, the circle is a unit circle the length of hypotenuse would be equal to 1.

Hypotenuse = 1

So, the formula of csc would be reduced to:

$csc(e) = \frac{1}{\text{Opposite}}$

The side opposite to the angle made at origin is y. So,

$csc(e) = \frac{1}{y}$

6 0

2 years ago

Read 2 more answers

Write the fraction 1 8 as a percent round to the nearest hundredth of a percent where necessary

Airida [17]

.125 is the answer to that question

7 0

2 years ago

5. Add (-6-22) + (-3 2 1).<br> A.-[903]<br> B-[303]<br> C-[603]<br> D.-9-3-5

Scorpion4ik [409]

So it (-6-22)+(321) ? Before I solve this problem?

6 0

2 years ago

Find the sum of the positive integers less than 200 which are not multiples of 4 and 7

taurus [48]

Answer:

$12942$ is the sum of positive integers between $1$ (inclusive) and $199$ (inclusive) that are not multiples of $4$ and not multiples $7$ .

Step-by-step explanation:

For an arithmetic series with:

$a_1$ as the first term,
$a_n$ as the last term, and
$d$ as the common difference,

there would be $\displaystyle \left(\frac{a_n - a_1}{d} + 1\right)$ terms, where as the sum would be $\displaystyle \frac{1}{2}\, \displaystyle \underbrace{\left(\frac{a_n - a_1}{d} + 1\right)}_\text{number of terms}\, (a_1 + a_n)$ .

Positive integers between $1$ (inclusive) and $199$ (inclusive) include:

$1,\, 2,\, \dots,\, 199$ .

The common difference of this arithmetic series is $1$ . There would be $(199 - 1) + 1 = 199$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times ((199 - 1) + 1) \times (1 + 199) = 19900 \end{aligned}$ .

Similarly, positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $4$ include:

$4,\, 8,\, \dots,\, 196$ .

The common difference of this arithmetic series is $4$ . There would be $(196 - 4) / 4 + 1 = 49$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 49 \times (4 + 196) = 4900 \end{aligned}$

Positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $7$ include:

$7,\, 14,\, \dots,\, 196$ .

The common difference of this arithmetic series is $7$ . There would be $(196 - 7) / 7 + 1 = 28$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 28 \times (7 + 196) = 2842 \end{aligned}$

Positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $28$ (integers that are both multiples of $4$ and multiples of $7$ ) include:

$28,\, 56,\, \dots,\, 196$ .

The common difference of this arithmetic series is $28$ . There would be $(196 - 28) / 28 + 1 = 7$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 7 \times (28 + 196) = 784 \end{aligned}$ .

The requested sum will be equal to:

the sum of all integers from $1$ to $199$ ,
minus the sum of all integer multiples of $4$ between $1\!$ and $199\!$ , and the sum integer multiples of $7$ between $1$ and $199$ ,
plus the sum of all integer multiples of $28$ between $1$ and $199$ - these numbers were subtracted twice in the previous step and should be added back to the sum once.

That is:

$19900 - 4900 - 2842 + 784 = 12942$ .

8 0

2 years ago

The distribution of scores on a recent test closely followed a Normal Distribution with a mean of 22 points and a standard devia

maksim [4K]

6.68% of the students scored at least 19 points on this test while the 85 percentile has a test score of 24.08 points

The z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:

$z=\frac{x-\mu}{\sigma} \\\\where\ \mu=mean.\sigma=standard\ deviation, x=raw\ score\\\\Given\ that\ mean(\mu)=22,standard \ deviation (\sigma)=2\\\\For\ x>19:\\\\z=\frac{22-19}{2}=1.5$

P(x > 19) = P(z > 1.5) = 1 - P(z < 1.5) = 1 - 0.9332 = 0.0668

6.68% of the students scored at least 19 points on this test

b) 85 percentile corresponds to a z score of 1.04

$z=\frac{x-\mu}{\sigma}\\\\1.04 =\frac{x-22}{2} \\\\2.08=x-22\\\\x=24.08$

The 85 percentile has a test score of 24.08

Find out more at: brainly.com/question/25012216

7 0

2 years ago