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patriot [66]
3 years ago
14

I need help pls its due in 40 minutes

Mathematics
1 answer:
mina [271]3 years ago
4 0

Answer:

I am attaching photo hold on

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A baby gains 11 pounds in its first year of life. The baby gained 4 1/4 pounds during the first four months and 3 1/2 pounds in
Ede4ka [16]

Answer:

3.25 pounds

Step-by-step explanation:

x + 4 (1/2) + 3 (1/2) = 11

x + 4.25 + 3.5 = 11

x + 7.75 = 11

x = 11 - 7.75

x = 3.25 pounds

8 0
3 years ago
If a picnic area is located at the midpoint between Sacramento and Oakland, find the distance to the picnic area from the road s
Oxana [17]
First, think this through. If you are 23 miles from Sacramento, and 110 miles from Oakland (I am assuming it is miles since you didn't specify *cough cough*) then the two cities are 133 miles apart (also assuming you are in between them)
23miles + 110miles = 133miles
so, we use the midpoint formula to find our position
\frac{distance}{ 2}
and get 66.5 miles from both Sacramento and Oakland.

Next, we already know the sign is 23 miles away from Sacramento so we can use that distance and our position to figure out how far away the sign is. We use absolute values because distance is always positive.
|our \: position - distance \: of \: sign|

so we get
|66.5miles - 23miles|  = 43.5miles
to double check, use the sign's distance from Oakland.
|66.5miles - 110miles|  = 43.5miles
so, we are 43.5 miles from the sign!

hope this helps

7 0
3 years ago
Chnstopher earned $142.20. He saved $24.50. He spent the rest on 2 new pairs
pogonyaev

Answer:

$58.25

Step-by-step explanation:

142.20 - 24.50 = 117.50

117.50 / 2 = 58.25

6 0
3 years ago
Read 2 more answers
How to answer the question
tatyana61 [14]
5/60 = 12
12 x 3 =36
J
4 0
3 years ago
Read 2 more answers
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
3 years ago
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