Which symbol makes the following TRUE?

Jet001 [13]

There are only 6 place values in 128,955.

If you would round it it could be:

130,000
129,000
128,960

3 0

2 years ago

The coordinates of which point are (–3, –6)? On a coordinate plane, point A is (negative 3, 6), point B is (negative 6, negative

julia-pushkina [17]

Answer:

Point c

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

It is estimated that 0.54 percent of the callers to the Customer Service department of Dell Inc. will receive a busy signal. Wha

stira [4]

Using the binomial distribution, it is found that there is a 0.8295 = 82.95% probability that at least 5 received a busy signal.

<h3>What is the binomial distribution formula?</h3>

The formula is:

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

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

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

In this problem:

0.54% of the calls receive a busy signal, hence p = 0.0054.

A sample of 1300 callers is taken, hence n = 1300.

The probability that at least 5 received a busy signal is given by:

$P(X \geq 5) = 1 - P(X < 5)$

In which:

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4).

Then:

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

$P(X = 0) = C_{1300,0}.(0.0054)^{0}.(0.9946)^{1300} = 0.0009$

$P(X = 1) = C_{1300,1}.(0.0054)^{1}.(0.9946)^{1299} = 0.0062$

$P(X = 2) = C_{1300,2}.(0.0054)^{2}.(0.9946)^{1298} = 0.0218$

$P(X = 3) = C_{1300,3}.(0.0054)^{3}.(0.9946)^{1297} = 0.0513$

$P(X = 4) = C_{1300,4}.(0.0054)^{4}.(0.9946)^{1296} = 0.0903$

Then:

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0009 + 0.0062 + 0.0218 + 0.0513 + 0.0903 = 0.1705.

$P(X \geq 5) = 1 - P(X < 5) = 1 - 0.1705 = 0.8295$

0.8295 = 82.95% probability that at least 5 received a busy signal.

More can be learned about the binomial distribution at brainly.com/question/24863377

#SPJ1

6 0

1 year ago

Please Help!!

Genrish500 [490]

Given

$a\sqrt{x+b}+c=d$

we have

$\sqrt{x+b}=\dfrac{d-c}{a}$

Squaring both sides, we have

$x+b=\dfrac{(d-c)^2}{a^2}$

And finally

$x=\dfrac{(d-c)^2}{a^2}-b$

Note that, when we square both sides, we have to assume that

$\dfrac{d-c}{a}>0$

because we're assuming that this fraction equals a square root, which is positive.

So, if that fraction is positive you'll actually have roots: choose

$a=1,\ b=0,\ c=2,\ d=6$

and you'll have

$\sqrt{x}+2=6 \iff \sqrt{x}=4 \iff x=16$

Which is a valid solution. If, instead, the fraction is negative, you'll have extraneous roots: choose

$a=1,\ b=0,\ c=10,\ d=4$

and you'll have

$\sqrt{x}+10=4 \iff \sqrt{x}=-6$

Squaring both sides (and here's the mistake!!) you'd have

$x=36$

which is not a solution for the equation, if we plug it in we have

$\sqrt{x}+10=4 \implies \sqrt{36}+10=4 \implies 6+10=4$

Which is clearly false

7 0

3 years ago

Lisa spends part of her year as a member of a gym. She then finds a better deal at another gym

Tema [17]

I’m having trouble understanding this question.?

5 0

2 years ago