X÷0.4=2 1/9 ÷3 1/6? What does x equal? Explain your answer thoroughly.

HELP HELP HELP!!!<br> i need help with math pls help me.

12345 [234]

Answer:

I dont even know this myself o-o

Step-by-step explanation:

:D

5 0

3 years ago

Help fast please and thank u!!!! i will give a medal and i will give brainliest! need 2 answers tho! explain it all!

12345 [234]

A) The median is 22 because it is in the middle
B) The range is 44 because it is the lowest (10) plus the highest (34)
C) The interquartile range is 36 because its the highest of the box (20) plus the lowest of the bow (16)

3 0

4 years ago

The probabilities of poor print quality given no printer problem, misaligned paper, high ink viscosity, or printer-head debris a

nadezda [96]

Answer and explanation:

Given : The probabilities of poor print quality given no printer problem, misaligned paper, high ink viscosity, or printer-head debris are 0, 0.3, 0.4, and 0.6, respectively.

The probabilities of no printer problem, misaligned paper, high ink viscosity, or printer-head debris are 0.8, 0.02, 0.08, and 0.1, respectively.

Let the event E denote the poor print quality.

Let the event A be the no printer problem i.e. P(A)=0.8

Let the event B be the misaligned paper i.e. P(B)=0.02

Let the event C be the high ink viscosity i.e. P(C)=0.08

Let the event D be the printer-head debris i.e. P(D)=0.1

and the probabilities of poor print quality given printers are

$P(E|A)=0,\ P(E|B)=0.3,\ P(E|C)=0.4,\ P(E|D)=0.6$

First we calculate the probability that print quality is poor,

$P(E)=P(A)P(E|A)+P(B)P(E|B)+P(C)P(E|C)+P(D)P(E|D)$

$P(E)=(0)(0.8)+(0.3)(0.02)+(0.4)(0.08)+(0.6)(0.1)$

$P(E)=0+0.006+0.032+0.06$

$P(E)=0.098$

a. Determine the probability of high ink viscosity given poor print quality.

$P(C|E)=\frac{P(E|C)P(C)}{P(E)}$

$P(C|E)=\frac{0.4\times 0.08}{0.098}$

$P(C|E)=\frac{0.032}{0.098}$

$P(C|E)=0.3265$

b. Given poor print quality, what problem is most likely?

Probability of no printer problem given poor quality is

$P(A|E)=\frac{P(E|A)P(A)}{P(E)}$

$P(A|E)=\frac{0\times 0.8}{0.098}$

$P(A|E)=\frac{0}{0.098}$

$P(A|E)=0$

Probability of misaligned paper given poor quality is

$P(B|E)=\frac{P(E|B)P(B)}{P(E)}$

$P(B|E)=\frac{0.3\times 0.02}{0.098}$

$P(B|E)=\frac{0.006}{0.098}$

$P(B|E)=0.0612$

Probability of printer-head debris given poor quality is

$P(D|E)=\frac{P(E|D)P(D)}{P(E)}$

$P(D|E)=\frac{0.6\times 0.1}{0.098}$

$P(D|E)=\frac{0.06}{0.098}$

$P(D|E)=0.6122$

From the above conditional probabilities,

The printer-head debris problem is most likely given that print quality is poor.

3 0

3 years ago

Read 2 more answers

Based on the polynomial remainder theorem, what is the value of the function when x=−5?

miss Akunina [59]

Answer:

5.

Step-by-step explanation:

f(x)=x^4+12x^3+30x^2−12x+70

The remainder theorem says that f(a) is the value of f(x) when f(x) is divided by (x - a).

So when a = -5 f(a) will be 5.

Answer is 5.

5 0

2 years ago

What is x and y math help

Readme [11.4K]

<h3>Answer:</h3>

20 cans of cola
10 cans of root beer

<h3>Step-by-step explanation:</h3>

x and y are whatever you want them to be.

It can be convenient for solving a problem like this to use x and y to represent <em>what the problem is asking for</em>: the number of cans of cola and the number of cans of root beer. It is also convenient (less confusing) to use those variable names in the same order that the nouns of the problem are named:

... x = # of cans of cola

... y = # of cans of root beer

Then the problem statement tells you ...

... x + y = 30 . . . . . . . 30 cans total were bought

... x = 2y . . . . . . . . . . the number of cans of cola is twice the number of cans of root beer

_____

This set of equations is nicely solved by substitution: use the second equation to substitute for x in the first.

... (2y) +y = 30 . . . . . put 2y where x was

... 3y = 30 . . . . . . . . collect terms

... y = 10 . . . . . . . . . divide by 3

... 2y = x = 20

<em>You're not done yet. You need to answer the question the problem asks.</em>

Jared bought 20 cans of cola and 10 cans of root beer.

_____

<em>Comment on x and y</em>

You customarily see x and y as the variables of a problem. Personally, I like to use variables that remind me what they stand for. In this problem, I might use "c" for cans of cola and "r" for cans of root beer. Then when I've found the solution, I know exactly how it relates to what the question is asking.

Always start by writing down what the variables stand for (as we did here). Sometimes, this is called <em>writing a Let statement</em>: <u>Let</u> x = number of colas; <u>let</u> y = number of root beers.

<em>Comment on problems of this type</em>

When a proportional relationship is given between the items in a sum (2 cola cans for every root beer can), it is often convenient to work the problem in terms of groups of items. Here, a group of 3 items can consist of 2 cola cans and 1 root beer can. Then 30 items will be 10 groups, so 10 root beers and 20 colas. The problem is solved even before you can name the variables.

Even when the relationship isn't exactly proportional, you can add or subtract the extras and still work the problem this way. Had we said colas numbered 3 more than twice as many root beers, we could have our groups of 3 total 27 (30 less the 3 extra), giving 9 root beers and 21 colas (3 + 2·9).

8 0

3 years ago