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Brilliant_brown [7]
3 years ago
7

Elena has a pet parakeet that weighs 6 when measured in one unit and 170 when measured in a different unit. Which measurement is

in ounces, and which is in grams? 6__________170______
Mathematics
1 answer:
alisha [4.7K]3 years ago
8 0

Given:

Elena has a pet parakeet that weighs 6 when measured in one unit and 170 when measured in a different unit.

6__________ = 170______

To find:

The correct units from grams and ounces for these blanks.

Solution:

We know that, 1 ounce is greater than 1 gram.

1 ounce = 28.3495 grams

Multiply both sides by 6.

6 ounces = 170.097 grams

Approximate the values to the nearest whole number.

6 ounces = 170 grams

Therefore, 6 ounces = 170 grams.

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What is the value of this expression when n=5?
34kurt

Answer: 19

Step-by-step explanation:

The answer is 19 because, if n=5 then we have to divide 45 by 5. 45 divided by 5 equals 9. and to make sure we can multiply 9 and 5 and that equals 45. So now its saying what is 9 plus 10 and that equals 19. Just to make sure we can subtract 19 by 10 or 9 and if we do eitheer one they both equal 19.

Hope This Helps!

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The answer to this problem is 2
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If you combine -30 and +75 that will be 40. so GCF will be 40

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You are at a stall at a fair where you have to throw a ball at a target. There are two versions of the game. In the first
Tomtit [17]

Answer:

P(X=0)=(3C0)(0.1)^0 (1-0.1)^{3-0}=0.729

And the probability of loss with the first wersion is 0.729

P(Y=0)=(5C0)(0.05)^0 (1-0.05)^{5-0}=0.774

And the probability of loss with the first wersion is 0.774

As we can see the best alternative is the first version since the probability of loss is lower than the probability of loss on version 2.

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

Alternative 1

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

X \sim Binom(n=3, p=0.1)

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!}

We can find the probability of loss like this P(X=0) and if we find this probability we got this:

P(X=0)=(3C0)(0.1)^0 (1-0.1)^{3-0}=0.729

And the probability of loss with the first wersion is 0.729

Alternative 2

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

Y \sim Binom(n=5, p=0.05)

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

P(Y)=(nCy)(p)^y (1-p)^{n-y}

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

nCy=\frac{n!}{(n-y)! y!}

We can find the probability of loss like this P(Y=0) and if we find this probability we got this:

P(Y=0)=(5C0)(0.05)^0 (1-0.05)^{5-0}=0.774

And the probability of loss with the first wersion is 0.774

As we can see the best alternative is the first version since the probability of loss is lower than the probability of loss on version 2.

4 0
4 years ago
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