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Norma-Jean [14]
3 years ago
8

A conference room is in the shape of a rectangle. Its floor has a length of (x − 4) meters and a width of (3x − 1) meters. The e

xpression below represents the area of the floor of the room in square meters:
(x − 4)(3x − 1)

Which of the following simplified expressions represents the area of the floor of the conference room in square meters?

x2 − 13x + 4
3x2 − 13x + 4
3x2 − 11x + 4
x2 − 12x + 4
Mathematics
2 answers:
Naddik [55]3 years ago
7 0

Answer:

3x^2−13x+4

Step-by-step explanation:

(x−4)(3x−1)

=(x+−4)(3x+−1)

=(x)(3x)+(x)(−1)+(−4)(3x)+(−4)(−1)

=3x^2−x−12x+4

=3x^2−13x+4

weqwewe [10]3 years ago
4 0

Answer:

Option A. 3x² - 13x + 4 is the answer.

Step-by-step explanation:

A conference room is in the shape of a rectangle. Its floor has a length of (x - 4) and width of (3x - 1) meters.

The expression that represents the area of the room is (x - 4)(3x -1) meter²

We further simplify this expression representing the area of the conference room.

(x - 4)(3x -1) = x(3x -1) - 4(3x -1) [Distributive law]

= 3x² - x - 12x + 4

= 3x² - 13x + 4

Option A. 3x² - 13x + 4 is the simplified form of the given expression.

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Answer:

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3 years ago
Suppose a geyser has a mean time between eruptions of 72 minutes. Let the interval of time between the eruptions be normally dis
nikitadnepr [17]

Answer:

(a) The probability that a randomly selected time interval between eruptions is longer than 82 ​minutes is 0.3336.

(b) The probability that a random sample of 13-time intervals between eruptions has a mean longer than 82 ​minutes is 0.0582.

(c) The probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 ​minutes is 0.0055.

(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.

(e) The population mean must be more than 72​, since the probability is so low.

Step-by-step explanation:

We are given that a geyser has a mean time between eruptions of 72 minutes.

Also, the interval of time between the eruptions be normally distributed with a standard deviation of 23 minutes.

(a) Let X = <u><em>the interval of time between the eruptions</em></u>

So, X ~ N(\mu=72, \sigma^{2} =23^{2})

The z-score probability distribution for the normal distribution is given by;

                            Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean time = 72 minutes

           \sigma = standard deviation = 23 minutes

Now, the probability that a randomly selected time interval between eruptions is longer than 82 ​minutes is given by = P(X > 82 min)

       P(X > 82 min) = P( \frac{X-\mu}{\sigma} > \frac{82-72}{23} ) = P(Z > 0.43) = 1 - P(Z \leq 0.43)

                                                           = 1 - 0.6664 = <u>0.3336</u>

The above probability is calculated by looking at the value of x = 0.43 in the z table which has an area of 0.6664.

(b) Let \bar X = <u><em>sample mean time between the eruptions</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time = 72 minutes

           \sigma = standard deviation = 23 minutes

           n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between eruptions has a mean longer than 82 ​minutes is given by = P(\bar X > 82 min)

       P(\bar X > 82 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{82-72}{\frac{23}{\sqrt{13} } } ) = P(Z > 1.57) = 1 - P(Z \leq 1.57)

                                                           = 1 - 0.9418 = <u>0.0582</u>

The above probability is calculated by looking at the value of x = 1.57 in the z table which has an area of 0.9418.

(c) Let \bar X = <u><em>sample mean time between the eruptions</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time = 72 minutes

           \sigma = standard deviation = 23 minutes

           n = sample of time intervals = 34

Now, the probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 ​minutes is given by = P(\bar X > 82 min)

       P(\bar X > 82 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{82-72}{\frac{23}{\sqrt{34} } } ) = P(Z > 2.54) = 1 - P(Z \leq 2.54)

                                                           = 1 - 0.9945 = <u>0.0055</u>

The above probability is calculated by looking at the value of x = 2.54 in the z table which has an area of 0.9945.

(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.

(e) If a random sample of 34-time intervals between eruptions has a mean longer than 82 ​minutes, then we conclude that the population mean must be more than 72​, since the probability is so low.

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3 years ago
Isolate the absolute value expression in the following equations to determine if they can be solved if so find and graph the sol
aev [14]

Answer:

x=-4

Step-by-step explanation:

2(|x+4|+3)=6

take out absolute value

2(x+4+3)=6

remove parrenthesis

2x+8+6=6

Simplify

2x+14=6

    -14  -14

2x=-8

x=-4

Hope this helps!!!!!

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3 years ago
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