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tensa zangetsu [6.8K]

3 years ago

11

Your car breaks down in a remote location. your cell phone has enough power left in it to send four text messages. if there is a

30% chance that any text you send will be read by the recipient, what is the probability that you will send all four texts in order to reach help (i.e., the probability that your fourth text will be the first and only one read)?

1 answer:

ale4655 [162]3 years ago

3 0

There is a 70% chance that any one text will not be read, so 0.7³×0.3=0.1029 is the probability that the fourth text will be read. This is the same as 10.29%.

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Answer asap please

Oduvanchick [21]

Answer:

(0,1) and (-2,9)

Step-by-step explanation:

A point (x,y) lies on the graph of f(x) = $(\frac{1}{3} )^{x}$ if it satisfies the condition y = $(\frac{1}{3} )^{x}$

<h3>(0,1):</h3>

x = 0 and y = 1.

This point satisfies the required condition as

$(\frac{1}{3} )^{0}$ = 1

Hence, this point <u>is</u> on the graph of f(x) = $(\frac{1}{3} )^{x}$

<h3>(3,27):</h3>

x = 3 and y = 27.

This point doesn't satisfy the required condition as

$(\frac{1}{3} )^{3}$ = $\frac{1}{27}$ ≠ 27

Hence, this point i<u>s not</u> on the graph of f(x) = $(\frac{1}{3} )^{x}$

<h3>(-2,9):</h3>

x = -2 and y = 9.

This point satisfies the required condition as

$(\frac{1}{3} )^{-2}$ = $3^{2}$ = 9

Hence, this point <u>is</u> on the graph of f(x) = $(\frac{1}{3} )^{x}$

<h3>(-1, $-\frac{1}{3}$ ):</h3>

x = -1 and y = $-\frac{1}{3}$ .

This point satisfies the required condition as

y = $(\frac{1}{3} )^{-1}$ = $3^{1}$ = 3 ≠ $-\frac{1}{3}$

Hence, this point <u>is not</u> on the graph of f(x) = $(\frac{1}{3} )^{x}$

7 0

3 years ago

What is equivalent to (9f + 12)*4)

Natalija [7]

Answer:

10

Step-by-step explanation:

3 0

3 years ago

Use synthetic division to evaluate (3x2 − 6x + 3) ÷ (x − 1). 3x − 9 3x2 − 3x 3x − 3 3x2 − 6

Viefleur [7K]

Answer:

$\frac{\left(3x^2-6x+3\right)}{\left(x-1\right)3x-93x^2-3x^3x-33x^2-6}: -\frac{\left(x-1\right)^2}{x^4+41x^2+x+2}$
$\frac{3x^2-6x+3}{\left(x-1\right)\cdot \:3x-126x^2-3x^3x-6}$
$\frac{3\left(x-1\right)^2}{\left(x-1\right)\cdot \:3x-126x^2-3x^3x-6}$
$\frac{3\left(x-1\right)^2}{-3\left(x^4+41x^2+x+2\right)}$
$\frac{\left(x-1\right)^2}{-\left(x^4+41x^2+x+2\right)}$
$-\frac{\left(x-1\right)^2}{x^4+41x^2+x+2}$

3 0

2 years ago

Factor completely.<br><br> -5x 2 y + 10xy - 15xy 2

slavikrds [6]

I think it is -30xy.

4 0

4 years ago

A search committee is formed to find a new software engineer.

Free_Kalibri [48]

Answer:

(a) 1,902,231,808,400

(b) 84

(c) 20

Step-by-step explanation:

In mathematics, the procedure to select k items from n distinct items, without replacement, is known as combinations.

The formula to compute the combinations of k items from n is given by the formula:

${n\choose k}=\frac{n!}{k!\cdot(n-k)!}$

(a)

Compute the number of ways to select 9 applicants from 100 as follows:

${100\choose 9}=\frac{100!}{9!\cdot(100-9)!}$

$=\frac{100!}{9!\times 91!}\\\\=\frac{100\times 99\times 98\times 97\times 96\times 95\times 94\times 93\times 92\times 91!}{9!\times 91!}\\\\=\frac{100\times 99\times 98\times 97\times 96\times 95\times 94\times 93\times 92}{9!}\\\\=1902231808400$

(b)

Compute the number of ways to select 6 people from 9 as follows:

${9\choose 6}=\frac{9!}{6!\cdot(9-6)!}$

$=\frac{9!}{6!\times 3!}\\\\=\frac{9\times 8\times 7\times 6!}{6!\times 3!}\\\\=\frac{9\times 8\times 7}{3!}\\\\=84$

(c)

Compute the number of ways to select top 3 candidates from 6 as follows:

${6\choose 3}=\frac{6!}{3!\cdot(6-3)!}$

$=\frac{6!}{3!\times 3!}\\\\=\frac{6\times 5\times 4\times 3!}{3!\times 3!}\\\\=\frac{6\times 5\times 4}{3!}\\\\=20$

7 0

3 years ago