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3 years ago

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On Naomi's cell phone plan, the amount she pays each month for international text messages is proportional to the number of inte

rnational texts she sends that month. Last month, she paid $3.20 for 16 international texts.
A. What is the constant of proportionality in this proportional relationship?
B. Write an equation to represent this proportional relationship. Make sure to define the variables you use.

1 answer:

Lostsunrise [7]3 years ago

8 0

A) The constant of proportionality in this proportional relationship is $k = \frac{y}{x}$

B) The equation to represent this proportional relationship is y = 0.2x

<h3><u>Solution:</u></h3>

Given that,

The amount Naomi pays each month for international text messages is proportional to the number of international texts she sends that month

Therefore,

This is a direct variation proportion

$\text{ amount Naomi pays each month } \propto \text{ number of international texts she sends that month}$

Let "y" be the amount that Naomi pays each month

Let "x" be the number of international texts she sends that month

Therefore,

$y \propto x$

y = kx -------- eqn 1

Where, "k" is the constant of proportionality

Thus the constant of proportionality in this proportional relationship is:

$k = \frac{y}{x}$

<em><u>Last month, she paid $3.20 for 16 international texts</u></em>

Therefore,

y = 3.20

x = 16

Thus from eqn 1,

$3.20 = k \times 16\\\\k = \frac{3.20}{16}\\\\k = 0.2$

Substitute k = 0.2 in eqn 1

y = 0.2x

The equation would then be y = 0.2x

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Read 2 more answers

When circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%. Let

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(a) P(X=3) = 0.093

(b) P(X≤3) = 0.966

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(e) The probability that none of the 25 boards is defective is 0.277.

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Step-by-step explanation:

We are given that when circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%.

Let X = <em>the number of defective boards in a random sample of size, n = 25</em>

So, X ∼ Bin(25,0.05)

The probability distribution for the binomial distribution is given by;

$P(X=r)= \binom{n}{r} \times p^{r}\times (1-p)^{n-r} ; x = 0,1,2,......$

where, n = number of trials (samples) taken = 25

r = number of success

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(a) P(X = 3) = $\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}$

= $2300 \times 0.05^{3}\times 0.95^{22}$

= <u>0.093</u>

(b) P(X $\leq$ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

= $\binom{25}{0} \times 0.05^{0}\times (1-0.05)^{25-0}+\binom{25}{1} \times 0.05^{1}\times (1-0.05)^{25-1}+\binom{25}{2} \times 0.05^{2}\times (1-0.05)^{25-2}+\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}$

= $1 \times 1 \times 0.95^{25}+25 \times 0.05^{1}\times 0.95^{24}+300 \times 0.05^{2}\times 0.95^{23}+2300 \times 0.05^{3}\times 0.95^{22}$

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= 1 - 0.966

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<u></u>

(d) P(1 ≤ X ≤ 3) = P(X = 1) + P(X = 2) + P(X = 3)

= $\binom{25}{1} \times 0.05^{1}\times (1-0.05)^{25-1}+\binom{25}{2} \times 0.05^{2}\times (1-0.05)^{25-2}+\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}$

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= $25 \times 0.05$ = 1.25

The standard deviation of X is given by;

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