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3 years ago

7

Samantha loves to download new music. She originally had 52 songs but plans to purchase 2 new songs each week. She wants to know

how many songs she will have after 93 weeks. Which equation should she use?
a93 = 93 + 2(52 − 1)
a93 = 93 + 2(52 + 1)
a93 = 52 + 2(93 + 1)
a93 = 52 + 2(93 − 1)

Hint: It's either A, B, C, or D

2 answers:

Nat2105 [25]3 years ago

6 0

$a_1=52\\
d=2\\
n=93\\\\
a_n=a_1+(n-1)\cdot d\\\\
a_{93}=52+(93-1)\cdot2=52+2(93-1)
$

natta225 [31]3 years ago

5 0

They want to construct this in the form of an arithmetic sequence which as the form:

a(n)=a+d(n-1), where a=initial value, d=common difference.

In this case a=52 and d=2, so our formula is:

a(n)=52+2(n-1) and if we are looking for the 93 term we have

a(93)=52+2(93-1)

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3 years ago

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Answer:

C. $f(x)=\sqrt[3]{-x} -1$

Step-by-step explanation:

Consider graph of the parent function (red curve in attached diagram)

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$h(x)=-\sqrt[3]{x}$

Then translate the graph of the function h(x) 1 unit down, then you'll get the function

$f(x)=-\sqrt[3]{x} -1\\ \\ \text{or}\\ \\f(x)=\sqrt[3]{-x} -1$

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3 years ago

A fair coin is to be tossed 20 times. Find the probability that 10 of the tosses will fall heads and 10 will fall tails, (a) usi

lbvjy [14]

Using the distributions, it is found that there is a:

a) 0.1762 = 17.62% probability that 10 of the tosses will fall heads and 10 will fall tails.

b) 0% probability that 10 of the tosses will fall heads and 10 will fall tails.

c) 0.1742 = 17.42% probability that 10 of the tosses will fall heads and 10 will fall tails.

Item a:

Binomial probability distribution

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

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

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:

20 tosses, hence $n = 20$ .
Fair coin, hence $p = 0.5$ .

The probability is <u>P(X = 10)</u>, thus:

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

$P(X = 10) = C_{20,10}.(0.5)^{10}.(0.5)^{10} = 0.1762$

0.1762 = 17.62% probability that 10 of the tosses will fall heads and 10 will fall tails.

Item b:

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.
After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
The binomial distribution is the probability of <u>x successes on n trials, with p probability</u> of a success on each trial. It can be approximated to the normal distribution with $\mu = np, \sigma = \sqrt{np(1-p)}$ .

The probability of an exact value is 0, hence 0% probability that 10 of the tosses will fall heads and 10 will fall tails.

Item c:

For the approximation, the mean and the standard deviation are:

$\mu = np = 20(0.5) = 10$

$\sigma = \sqrt{np(1 - p)} = \sqrt{20(0.5)(0.5)} = \sqrt{5}$

Using continuity correction, this probability is $P(10 - 0.5 \leq X \leq 10 + 0.5) = P(9.5 \leq X \leq 10.5)$ , which is the <u>p-value of Z when X = 10.5 subtracted by the p-value of Z when X = 9.5.</u>

X = 10.5:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{10.5 - 10}{\sqrt{5}}$

$Z = 0.22$

$Z = 0.22$ has a p-value of 0.5871.

X = 9.5:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{9.5 - 10}{\sqrt{5}}$

$Z = -0.22$

$Z = -0.22$ has a p-value of 0.4129.

0.5871 - 0.4129 = 0.1742.

0.1742 = 17.42% probability that 10 of the tosses will fall heads and 10 will fall tails.

A similar problem is given at brainly.com/question/24261244

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3 years ago

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I think this might be the answer since (a+b) equals 4, and ab= 3. I'm not sure.
a3 + b3
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= 12+12
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3 years ago