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9966 [12]
3 years ago
9

Solve the equation 425x=850

Mathematics
1 answer:
mylen [45]3 years ago
7 0

Answer:2


Step-by-step explanation:

DIVIDE 425 ON BOTH SIDES

AND U WILL GET 2

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Allison’s cupcake and cookie budget for her party is $105. She would like to have at least 40 cupcakes and cookies combined, and
FinnZ [79.3K]

Answer:

a + b ≥ 40

a + 5 ≤ b

2.50a + 1.50b ≤ 105

Step-by-step explanation:

We are told she would like to have at least 40 cupcakes and cookies combined.

If a is number of cupcakes and b is number of cookies, then this inequality is;

a + b ≥ 40

Secondly, we are told she would like to have at most 5 more cupcakes than cookies.

Thus, the inequality is;

a + 5 ≤ b

Lastly, we are told that cupcakes sell for $2.50 and cookies for $1.50

Thus, the inequality is;

2.50a + 1.50b ≤ 105

Finally, the 3 inequalities are;

a + b ≥ 40

a + 5 ≤ b

2.50a + 1.50b ≤ 105

4 0
3 years ago
2) What x value makes the exponential function greater than the
Ludmilka [50]

Answer:

The answer is b

Step-by-step explanation:

5 0
2 years ago
We learned in Exercise 4.18 that about 90% of American adults had chickenpox before adulthood. We now consider a random sample o
liraira [26]

Answer:

a) 108 people with a standard deviation of 3.286335

b) No

c) 0.218163 or around 21.82%

d) See explanation below.

Step-by-step explanation:

This situation can be modeled with the Binomial Distribution which gives t<em>he probability of an event that occurs exactly k times out of n, and is given by </em>

<em>\large P(k;n)=\binom{n}{k}p^kq^{n-k} </em>

<em>where  </em>

<em>\large \binom{n}{k}= combination of n elements taken k at a time. </em>

<em>p = probability that the event (“success”) occurs once </em>

<em>q = 1-p </em>

In this case, the event “success” is finding an American adult who had  chickenpox before adulthood with probability p=0.9

and n=120 American adults in the sample.

The <em>standard deviation for this binomial distribution</em> is

\large s=\sqrt{npq}

where <em>n is the sample size 120 </em>

a)

We consider a random sample of 120 American adults. How many people in this sample would you expect to have had chickenpox in their childhood?

If about 90% of American adults had chickenpox before adulthood, we expect to find 90% of 120 = 0.9*120=108 people in the sample who had chickenpox in their childhood.

The  standard deviation would be

\large s=\sqrt{120*0.9*0.1=3.286335}

b)

Would you be surprised if there were 105 people who have had chickenpox in their childhood?

No, because 105 and 108 are in the interval [mean - s, mean +s]

c)

What is the probability that 105 or fewer people in this sample have had chickenpox in their childhood?

The probability that 105 or fewer people in this sample have had chickenpox in their childhood is

\large \sum_{k=0}^{105}\binom{120}{k}0.9^k0.1^{120-k}

We compute this value easily with a spreadsheet and we get

\large \sum_{k=0}^{105}\binom{120}{k}0.9^k0.1^{120-k}=0.218163\approx 21.82\%

d)

How does this probability relate to your answer to part (b)?

A binomial distribution with np>5 and nq>5 with n the sample size as is the case here, behaves pretty much like a Normal distribution with mean np and standard deviation \large \sqrt{npq}, so around 60% of the data are in the interval  [mean -s, mean +s] and 40% outside, so roughly <em>20% of the data should be in [0, mean-s] </em>

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Answer:

110000=p(1+0.019)^168

23.62p=110000

p=4660

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3 years ago
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