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2 years ago

Is 1.150 and 1.15 the same? (ANSWER QUICKLY)

Mathematics

1 answer:

Ann [662]2 years ago

3 0

Yes because the 0 at the end is just a place holder basically so 1.150 and 1.15 are the same

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1-1: HOMEWORK

Paraphin [41]

I think its 0.78/1...???

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2 years ago

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Gabriel works at an amusement park. He loads 15 new passengers onto the roller coaster every 5/6 of a minute. At what rate does

dem82 [27]

Answer:

12 1/2 or 12.5

Step-by-step explanation:

all you have to do is multiply 15 times 5/6

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2 years ago

Write an expression for each phrase

Svetradugi [14.3K]

1. 9 + k - 19

2. 6.7n

3. 37t - 9.85

5. 15 + $\frac{60}{w}$

6. 7 - 3v

7. $5m - \frac{t}{7}$

8. $\frac{p}{14} + \frac{q}{3}$

9. 8 - 9r

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2 years ago

4. Use SUBSTITUTION to solve the system. x= y + 5 2y + x = -4

Marianna [84]

Answer:

if I I just want free point so I am

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3 years ago

A real estate agent has 19 properties that she shows. She feels that there is a 30% chance of selling any one property during a

netineya [11]

Answer:

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

We can find the individual probabilities:

$P(X=0)=(19C0)(0.3)^0 (1-0.3)^{19-0}=0.00114$

$P(X=1)=(19C1)(0.3)^1 (1-0.3)^{19-1}=0.0092$

$P(X=2)=(19C2)(0.3)^2 (1-0.3)^{19-2}=0.0358$

$P(X=3)=(19C3)(0.3)^3 (1-0.3)^{19-3}=0.0869$

$P(X=4)=(19C4)(0.3)^4 (1-0.3)^{19-4}=0.1491$

And replacing we got:

$P(X \geq 5) = 1-[0.00114+0.009282+0.0358+0.0869+0.149]= 0.7178$

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:

$X \sim Binom(n=19, p=0.3)$

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:

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

And we want to find this probability:

$P(X \geq 5)$

And we can use the complement rule:

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

We can find the individual probabilities:

$P(X=0)=(19C0)(0.3)^0 (1-0.3)^{19-0}=0.00114$

$P(X=1)=(19C1)(0.3)^1 (1-0.3)^{19-1}=0.0092$

$P(X=2)=(19C2)(0.3)^2 (1-0.3)^{19-2}=0.0358$

$P(X=3)=(19C3)(0.3)^3 (1-0.3)^{19-3}=0.0869$

$P(X=4)=(19C4)(0.3)^4 (1-0.3)^{19-4}=0.1491$

And replacing we got:

$P(X \geq 5) = 1-[0.00114+0.009282+0.0358+0.0869+0.149]= 0.7178$

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3 years ago