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1 year ago

HELP ASAP LIKE RIGHT NOW

Mathematics

2 answers:

leva [86]1 year ago

8 0

The answer is G. 12.6

dlinn [17]1 year ago

6 0

My best guess is add all of the numbers together then divide by 6 then check your choices to see if your response matches

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Please can I have an explanation also, I am terrible at these kinds of questions!

wlad13 [49]

Answer:

The fraction of the beads that are red is

Step-by-step explanation:

Algebraic Expressions

A bag contains red (r), yellow (y), and blue (b) beads. We are given the following ratios:

r:y = 2:3

y:b = 5:4

We are required to find r:s, where s is the total of beads in the bag, or

s = r + y + b

Thus, we need to calculate:

$\displaystyle \frac{r}{r+y+b} \qquad\qquad [1]$

Knowing that:

$\displaystyle \frac{r}{y}=\frac{2}{3} \qquad\qquad [2]$

$\displaystyle \frac{y}{b}=\frac{5}{4}$

Multiplying the equations above:

$\displaystyle \frac{r}{y}\frac{y}{b}=\frac{2}{3}\frac{5}{4}$

Simplifying:

$\displaystyle \frac{r}{b}=\frac{5}{6} \qquad\qquad [3]$

Dividing [1] by r:

$\displaystyle \frac{r}{r+y+b}=\displaystyle \frac{1}{1+y/r+b/r}$

Substituting from [2] and [3]:

$\displaystyle \frac{r}{r+y+b}=\displaystyle \frac{1}{1+3/2+6/5}$

Operating:

$\displaystyle \frac{r}{r+y+b}=\displaystyle \frac{1}{\frac{10+3*5+6*2}{10}}$

$\displaystyle \frac{r}{r+y+b}=\displaystyle \frac{10}{10+15+12}$

$\displaystyle \frac{r}{r+y+b}=\displaystyle \frac{10}{37}$

The fraction of the beads that are red is $\mathbf{\frac{10}{37}}$

8 0

2 years ago

In the diagram below of parallelogram ROCK, m angle is 70 degrees and m angle ROS is 65 what is the value of x

il63 [147K]

Answer:

135

Step-by-step explanation:

5 0

2 years ago

A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 stude

EleoNora [17]

Answer:

a) $P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)$

And we can use the probability mass function and we got:

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

And adding we got:

$P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061$

b) $P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182$

c) $P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

$P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054$

And replacing we got:

$P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886$

d) $E(X) = 20*0.2= 4$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=20, p=0.2)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

We want this probability:

$P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)$

And we can use the probability mass function and we got:

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

And adding we got:

$P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061$

Part b

We want this probability:

$P(X=4)$

And using the probability mass function we got:

$P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182$

Part c

We want this probability:

$P(X>3)$

We can use the complement rule and we got:

$P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

$P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054$

And replacing we got:

$P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886$

Part d

The expected value is given by:

$E(X) = np$

And replacing we got:

$E(X) = 20*0.2= 4$

3 0

2 years ago

16p=208 Plz answer and I'll give you brainliest.

Leno4ka [110]

Answer:

p = 13

Step-by-step explanation:

Step 1: Write equation

16p = 208

Step 2: Solve for p

Divide both sides by 16: p = 13

Step 3: Check

Plug in p to verify it's a solution.

16(13) = 208

208 = 208

8 0

2 years ago

Read 2 more answers

Help please?

Alla [95]

$\bf \begin{array}{clclll}
3^{1001}&\qquad &4^{1002}\\
\uparrow &&\uparrow \\
a&&b
\end{array}\\\\
-----------------------------\\\\
(a+b)^2-(a-b)^2\implies (a^2+2ab+b^2)-(a^2-2ab+b^2)
\\\\\\
a^2+2ab+b^2-a^2+2ab-b^2\implies 2ab+2ab\implies 4ab
\\\\\\
now\qquad 4(3^{1001})(4^{1002})=k12^{1001}\qquad 
\begin{cases}
12^{1001}\\
(4\cdot 3)^{1001}\\
4^{1001}\cdot 3^{1001}
\end{cases}$

$\bf \\\\\\
4(3^{1001})(4^{1002})=k(4^{1001}\cdot 3^{1001})\implies \cfrac{4(3^{1001})(4^{1002})}{4^{1001}\cdot 3^{1001}}=k
\\\\\\
4\cdot \cfrac{3^{1001}}{3^{1001}}\cdot \cfrac{4^{1002}}{4^{1001}}=k\implies 4\cdot 4^{1002}4^{-1001}=k\impliedby 
\begin{array}{llll}
\textit{same base}\\
\textit{add the}\\
exponents
\end{array}
\\\\\\
4\cdot 4^{1002-1001}=k\implies 4\cdot 4^1=k\implies 16=k$

6 0

2 years ago