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lozanna [386]
2 years ago
15

Describe Sam's location relative to the theater

Mathematics
2 answers:
AVprozaik [17]2 years ago
7 0
What do you mean ? Is there a problem
sveticcg [70]2 years ago
3 0
Can you be a little more specific. Like where Sam lives and theater location.
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Which terms could be used as the first term of the expression below to create a polynomial written in standard form? Select five
kondaur [170]

Answer:

The last two terms of the expression are

  ----+8r^2s^4-3r^3s^3

Both the last terms has variable of degree equal to (2+4=6) and (3+3=6).So, the first term must have degree greater than 6.

Correct Options are

 1.\rightarrow 3r^4s^5\\\\2.\rightarrow -r^4s^6

3 0
2 years ago
Beatrice graphed the relationship between the
sdas [7]

Answer:

Option C. The time in seconds that passed before the printer started printing pages

see the explanation

Step-by-step explanation:

Let

y ---->the number of pages printed.

x --->  the  time (in seconds) since she sent a print job to  the printer

we know that

The x-intercept is the value of x when the value of y is equal to zero

In the context of the problem

The x-intercept is the time in seconds that passed before the printer started printing pages (the number of pages printed is equal to zero)

5 0
2 years ago
Read 2 more answers
An expression is given: 2 open parentheses square root of k minus 1 close parenthesis plus square root of 8. If on adding negati
neonofarm [45]

Answer:

Possible value of k is √2

Step-by-step explanation:

The information given are;

The expression, 2·(√k - 1) + √8 to which may be added -6·√2 to obtain a rational number, we therefore have;

2·(√k - 1) + √8 - 6·√2 = R

Therefore, simplifying gives;

2·√k - 2 + 2·√2  - 6·√2 =  2·√k - 2 - 4·√2 = R

2·√k - 2 - 4·√2 + 2= R + 2 = R

2·√k - 2+ 2 - 4·√2 = R

2·√k - 2+ 2 - 4·√2 = R

2·√k + 0 - 4·√2 = 2·√k - 4·√2 = 2·(√k - 2·√2) = R

(√k - 2·√2) = R/2 = R

Therefore, √2 is a factor of √k such that √k - 2·√2 = R

Which gives k = x·√2, where x = a rational number

When x = 1, k = √2.

Therefore, a possible value of k is √2

3 0
2 years ago
Paco's cell phone carrier charges him $0.20 for each text message he sends or receives, $0.15 per minute for calls, and a $15 mo
Katarina [22]

Given:

Paco's cell phone carrier charges him $0.20 for each text message he sends or receives, $0.15 per minute for calls, and a $15 monthly service fee. Paco is trying to keep his bill for the month below $30.

To Find:

The number of texts 't' Paco can send/receive in a month.

Answer:

0\leq t

best describes the number of texts he can send or receive to keep his bill less than $30 in a month.

Step-by-step explanation:

Paco wants to keep is monthly bill below $30.

We see that he has to pay a foxed monthly service fee of $15. This means he is only left with a limit of $30 - $15 = $15 for his monthly calls and texts.

That is, the amount he has to pay for texting and calling has to be less than $15.

For texts, the cell phone carrier charges $0.20 for sending/receiving texts.

For calls, he is charged $0.15 per minute.

Let the number of text messages Paco can send or receive in a month be denoted by 't'.

Let the number of minutes Paco can call in a month be denoted by 'c'.

Then, the total cost of text messages he can send or receive per month would be 0.20t and the total cost of the minutes he spends on calls would be 0.15c. Together, the sum of these has to be less than $15 if his monthly bill has to be kept less than $30 (accounting for the monthly service fee).

So,

0.20t+0.15c

The number of texts he can send will dpend on the number of minutes he spends on his calls. For Paco to spend maximum number of texts, he has to spend 0 minutes on calls.

So, putting c = 0, the aboce equation can be written as

0.20t+0

That is, Paco has to send and receive less than 75 texts.

So,

0\leq t

best describes the number of texts he can send or receive to keep his bill less than $30 in a month.

3 0
2 years ago
Read 2 more answers
(Ross 5.15) If X is a normal random variable with parameters µ " 10 and σ 2 " 36, compute (a) PpX ą 5q (b) Pp4 ă X ă 16q (c) PpX
pochemuha

Answer:

(a) 0.7967

(b) 0.6826

(c) 0.3707

(d) 0.9525

(e) 0.1587

Step-by-step explanation:

The random variable <em>X</em> follows a Normal distribution with mean <em>μ</em> = 10 and  variance <em>σ</em>² = 36.

(a)

Compute the value of P (X > 5) as follows:

P(X>5)=P(\frac{x-\mu}{\sigma}>\frac{5-10}{\sqrt{36}})\\=P(Z>-0.833)\\=P(Z

Thus, the value of P (X > 5) is 0.7967.

(b)

Compute the value of P (4 < X < 16) as follows:

P(4

Thus, the value of P (4 < X < 16) is 0.6826.

(c)

Compute the value of P (X < 8) as follows:

P(X

Thus, the value of P (X < 8) is 0.3707.

(d)

Compute the value of P (X < 20) as follows:

P(X

Thus, the value of P (X < 20) is 0.9525.

(e)

Compute the value of P (X > 16) as follows:

P(X>16)=P(\frac{x-\mu}{\sigma}>\frac{16-10}{\sqrt{36}})\\=P(Z>1)\\=1-P(Z

Thus, the value of P (X > 16) is 0.1587.

**Use a <em>z</em>-table for the probabilities.

8 0
2 years ago
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