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FromTheMoon [43]

3 years ago

10

A random survey shows that 14 out of 21 students plan to vote for Jacob for class president. Which is the best prediction of the

total number of votes Jacob will receive if 180 students vote?
A. 35
B. 90
C. 120
D.173

1 answer:

FromTheMoon [43]3 years ago

6 0

Answer: C. 120

Step-by-step explanation:

14/21 will get you 66.6. Taking that number and multiplying it by the 180 total students will end you with 120 which is your answer.

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Which graph has a slope of 4/5 ?

svlad2 [7]

Answer:

its B

Step-by-step explanation:

Its B

5 0

3 years ago

What conjecture can you make about the sum of the first 22 positive even numbers?

alekssr [168]

Answer:

A.) The sum of the product of the number of terms and the number of terms plus one.

Step-by-step explanation:

So, we have the sum of the first 22 positive even numbers:

S = 2 + 4+ 6 + ... + 44

We know the sum of such sequence is given by the (sum of the first number (2) and the last number (44)) multiplied by the number of terms (11).

S = (2 + 44) * 11 = 46 * 11 = 506

Answer A tells us it should be the sum of the number of terms (22) by the number of terms plus one (22 + 1 = 23).

S = 22 * 23 = 506

That's confirmed.

Answer B would give 22 * 21 = 462, so it's not right

Answer C would give 22 * 2 = 44, obviously not right

Answer D would give 22 * 22 = 484, close but not right.

7 0

2 years ago

Prove the following DeMorgan's laws: if LaTeX: XX, LaTeX: AA and LaTeX: BB are sets and LaTeX: \{A_i: i\in I\} {Ai:i∈I} is a fam

MariettaO [177]

$X-(A\cup B)=(X-A)\cap(X-B)$

I'll assume the usual definition of set difference, $X-A=\{x\in X,x\not\in A\}$ .

Let $x\in X-(A\cup B)$ . Then $x\in X$ and $x\not\in(A\cup B)$ . If $x\not\in(A\cup B)$ , then $x\not\in A$ and $x\not\in B$ . This means $x\in X,x\not\in A$ and $x\in X,x\not\in B$ , so it follows that $x\in(X-A)\cap(X-B)$ . Hence $X-(A\cup B)\subset(X-A)\cap(X-B)$ .

Now let $x\in(X-A)\cap(X-B)$ . Then $x\in X-A$ and $x\in X-B$ . By definition of set difference, $x\in X,x\not\in A$ and $x\in X,x\not\in B$ . Since $x\not A,x\not\in B$ , we have $x\not\in(A\cup B)$ , and so $x\in X-(A\cup B)$ . Hence $(X-A)\cap(X-B)\subset X-(A\cup B)$ .

The two sets are subsets of one another, so they must be equal.

$X-\left(\bigcup\limits_{i\in I}A_i\right)=\bigcap\limits_{i\in I}(X-A_i)$

The proof of this is the same as above, you just have to indicate that membership, of lack thereof, holds for all indices $i\in I$ .

Proof of one direction for example:

Let $x\in X-\left(\bigcup\limits_{i\in I}A_i\right)$ . Then $x\in X$ and $x\not\in\bigcup\limits_{i\in I}A_i$ , which in turn means $x\not\in A_i$ for all $i\in I$ . This means $x\in X,x\not\in A_{i_1}$ , and $x\in X,x\not\in A_{i_2}$ , and so on, where $\{i_1,i_2,\ldots\}\subset I$ , for all $i\in I$ . This means $x\in X-A_{i_1}$ , and $x\in X-A_{i_2}$ , and so on, so $x\in\bigcap\limits_{i\in I}(X-A_i)$ . Hence $X-\left(\bigcup\limits_{i\in I}A_i\right)\subset\bigcap\limits_{i\in I}(X-A_i)$ .

4 0

2 years ago

Write each of the following expressions without using absolute value: |z−6|−|z−5|, if z<5

abruzzese [7]

Answer:

The answer is <em>1</em>.

Step-by-step explanation:

Given the expression:

$|z-6|-|z-5|,\ if\ z$

To find:

The expression without absolute value.

Solution:

First of all, let us learn about the absolute value function:

$y = f(x) = |x| =\left \{ {{x\ if\ x>0} \atop {-x\ if\ x$

i.e. value is x if x is positive

value is -x if x is negative

Here the given expression contains two absolute value functions:

$|z-6|$ and $|z-5|$

Using the definition of absolute value function as per above definition.

$|z-5| =\left \{ {{(z-5)\ if\ z>5} \atop {-(z-5)\ if\ z$

$|z-6| =\left \{ {{(z-6)\ if\ z>6} \atop {-(z-6)\ if\ z$

Now, it is given that z < 5 that means z will also be lesser than 6 i.e. z < 6

So, given expression $|z-6|-|z-5|,\ if\ z$ will be equivalent to :

$-(z-6) - (-(z-5))\\\Rightarrow -z+6 +z-5 = \bold{1}$

So, the expression is equivalent to <em>1</em>.

6 0

3 years ago

Kenzie finds purple, blue, and silver space rocks while exploring, 2/3 of the rocks are purple and 3/4 of the remainder are blue

Nadya [2.5K]

Answer:

There were 36 space rocks in total.

Step-by-step explanation:

First, you need to find what fraction of the original rocks are silver.

If 2/3 are purple, then 1/3 of them are not.

If 3/4 of the remaining 1/3 are blue, then 3/4 x 1/3 = 3/12 rocks are blue.

2/3 of the rock are purple, 3/12 of them are blue, and the remaining is silver.

So, you need to do 2/3 - 3/12, then find common denominators so, 8/12 - 3/12 which = 1/12.

This means 1/12 of the rocks are silver. If there are silver rocks, you need to find a suitable equation that needs to be solved.

So:

1/12r = 3

r = 36 silver rocks

There were 36 space rocks in total.

hope this helps :)

3 0

2 years ago

Read 2 more answers