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3 years ago

What do you call a fraction with her at normal numerator and a denominator as 0 ?

2 answers:

7 0

Its basically called null, it's not anything, nothing can be divided by 0, although 0 can be divided by anything. So the answer is that it's not a fraction.

zzz [600]3 years ago

3 0

The answer is undefined because nothing can be over 0

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At an end of the year sale, Gabriela bought more than 12 bottles of hand soaps and lotions. If x represents the number of hand s

kolezko [41]

B.) x<span> + </span>y<span> > 12 should be the answer. </span>

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4 years ago

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Suppose an annuity pays 6% annual interest, compounded semi-annually. You invest in this annuity by contributing $4,500 semiannu

yaroslaw [1]

Answer:

$3240

Step-by-step explanation:

hope it is well understood

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3 years ago

Ps help i dont know maths

satela [25.4K]

Answer:

3526

Step-by-step explanation:

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3 years ago

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At a charity fund raiser, adult tickets were sold for $11 each and children’s tickets were sold for $5 each. Write an algebraic

patriot [66]

9514 1404 393

Answer:

11a +5c
$4667

Step-by-step explanation:

If 'a' adult tickets are sold for $11 each, the amount of money they sold for is ...

11a

If 'c' children's tickets are sold for $5 each, the amount of money they sold for is ...

Then the expression for the total amount from the sale of tickets is ...

11a +5c . . . . total of money raised

The value of this expression for a=247 and c=390 is ...

11·247 +5·390 = 2717 +1950 = 4667

A total of $4667 was raised by the sale of tickets.

3 0

3 years ago

The on-line access computer service industry is growing at an extraordinary rate. Current estimates suggest that only 20% of the

vitfil [10]

Answer:

The probability is 0.4207

Step-by-step explanation:

The probability of a home-based computer having access to on-line services is p = 0.2 (data from the exercise)

Then, the probability of a home-based computer not having access to on-line services is p = 1 - 0.2 = 0.8

We are going to use this probability (p = 0.8) to solve the exercise.

Let's define the random variable X

X : ''Number of home-based computers not having access to on-line services''

X can be modeled as a binomial random variable

X ~ Bi(p,n)

X ~Bi(0.8,25)

Where p is the success probability and n is the number of Bernoulli independent experiments we are taking place.

We are going to count ''a success'' as a computer not having access to on-line services.

The binomial probability function is :

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

Where P(X=x) is the probability of the random variable X to assume the value x

nCx is the combinatorial number define as

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

p is the success probability and n the number of Bernoulli independent experiments taking place.

In our exercise,

$p=0.8\\n=25$

We are looking for :

$P(X>20)=P(X=21)+P(X=22)+P(X=23)+P(X=24)+P(X=25)$

$P(X>20)=(25C21)0.8^{21}0.2^{4}+(25C22)0.8^{22}0.2^{3}+(25C23)0.8^{23}0.2^{2}+(25C24)0.8^{24}0.2^{1}+(25C25)0.8^{25}0.2^{0}$

$P(X>20)=0.1867+0.1358+0.0708+0.0236+0.8^{25}$

$P(X>20)=0.4207$

Finally, the probability of finding that more than 20 of 25 home-based computers do not have access to on-line services is 0.4207

6 0

3 years ago