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3 years ago

The set of life spans of an appliance is normally distributed with a mean

Mathematics

2 answers:

qaws [65]3 years ago

8 0

Answer:

24 months

Step-by-step explanation:

HACTEHA [7]3 years ago

5 0

Answer:

24 months

Step-by-step explanation:

Lifespan with z-score of -3 in the the set of normally distributed life spans of an appliance can be calculated using the equation:

$z=\frac{X-M}{s}$ where

X is the lifespan we are looking for

M is the mean life span of the appliance (48 months)

s is the standard deviation of the distribution of the life spans (8 months)

We have $-3=\frac{X-48}{8}$ that is

-24=X-48 and

X=24 months.

The life span of an appliance that has a z-score of -3 is 24 months.

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5 of what percent is 3.5

UNO [17]

If you take 3.5 and divide it by 0.05, or times 20, that would make it 70.

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3 years ago

Read 2 more answers

P is inversely proportional to the cube of (q-2) p=6 when q=3 find the value of p when q is 5

sveticcg [70]

$\bf \begin{array}{llllll}
\textit{something}&&\textit{varies inversely to}&\textit{something else}\\ \quad \\
\textit{something}&=&\cfrac{{{\textit{some value}}}}{}&\cfrac{}{\textit{something else}}\\ \quad \\
y&=&\cfrac{{{\textit{k}}}}{}&\cfrac{}{x}
\\
&&y=\cfrac{{{ k}}}{x}
\end{array}\\\\
-----------------------------\\\\
\textit{p is inversely proportional to the cube of (q-2)}\implies p=\cfrac{k}{(q-2)^3}
\\\\\\
now \quad 
\begin{cases}
p=6\\
q=3
\end{cases}\implies 6=\cfrac{k}{(3-2)^3}$

solve for "k", to find k or the "constant of variation"

then plug k's value back to $\bf p=\cfrac{k}{(q-2)^3}$

now.... what is "p" when q = 5? well, just set "q" to 5 on the right-hand-side, and simplify, to see what "p" is

4 0

3 years ago

What are the leading coefficient and degree of the polynomial?

MissTica

Answer:

leading coefficient: 2
degree: 7

Step-by-step explanation:

The degree of a term with one variable is the exponent of the variable. The degrees of the terms (in the same order) are ...

6, 0, 7, 1

The highest-degree term is 2x^7. Its coefficient is the "leading" coefficient, because it appears first when the polynomial terms are written in decreasing order of their degree:

2x^7 -7x^6 -18x -4

The leading coefficient is 2; the degree is 7.

<em>Additional comment</em>

When a term has more than one variable, its degree is the sum of the exponents of the variables. The term xy, for example, is degree 2.

4 0

2 years ago

34% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and a

finlep [7]

Answer:

a) There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

b) There is a 71.62% probability that more than two students use credit cards because of the rewards program.

c) There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

Step-by-step explanation:

There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this problem by the binomial distribution.

Binomial probability

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

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

And $\pi$ is the probability of X happening.

In this problem, we have that:

10 student are sampled, so $n = 10$

34% of college students say they use credit cards because of the rewards program, so $\pi = 0.34$

(a) exactly two

This is P(X = 2).

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

(b) more than two

This is $P(X > 2)$ .

Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities must be 1. So:

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

$P(X > 2) = 1 - P(X \leq 2)$

In which

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

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 0) = C_{10,0}.(0.34)^{0}.(0.66)^{10} = 0.0157$

$P(X = 1) = C_{10,1}.(0.34)^{1}.(0.66)^{9} = 0.0808$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0157 + 0.0808 + 0.1873 = 0.2838$

$P(X > 2) = 1 - P(X \leq 2) = 1 - 0.2838 = 0.7162$

There is a 71.62% probability that more than two students use credit cards because of the rewards program.

(c) between two and five inclusive

This is:

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X = 3) = C_{10,3}.(0.34)^{3}.(0.66)^{7} = 0.2573$

$P(X = 4) = C_{10,4}.(0.34)^{4}.(0.66)^{6} = 0.2320$

$P(X = 5) = C_{10,5}.(0.34)^{5}.(0.66)^{5} = 0.1434$

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1873 + 0.2573 + 0.2320 + 0.1434 = 0.82$

There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

6 0

3 years ago

Please helppp!!!! I’ll give you a crown!!

siniylev [52]

Answer:

the second one

Step-by-step explanation:

the last number of the equation (-3) is the y intercept, so it goes the the point (0,-3), also it has the opposite slope of y=2x-3

i hope this helps and im sorry if im wrong :(

8 0

3 years ago