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3 years ago

15 MINUTES LEFT PLEASE!

Mathematics

2 answers:

inysia [295]3 years ago

8 0

Answer:

65x^2 – 26x

Step-by-step explanation:

To determine the total number of seats, take the number of rows and multiply by the number of seats per row

f(x) * g(x)

13x * (5x-2)

65x^2 -26x

kow [346]3 years ago

3 0

Answer:

15 MINUTES LEFT PLEASE!

A wedding planner is organizing the seating for a wedding. He can represent the number of rows by the function f(x) = 13x

and the number of seats in each row by the function g(x) = 5x-2

Which function represents the total number of seats?

65x + 26

65x – 26

65x2 + 26x

65x2 – 26x✔

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A rental car company is running two specials. Customers can pay $50 to rent a compact car for the first day plus $6 for each add

natali 33 [55]

The number of additional days she wants to rent the car for the two specials are equivalent is 5 days.

<h3>Equation</h3>

Compact car:

Fixed price = $50
Additional price per day = $6

y = 50 + 6x

Another special:

Fixed price = $40
Additional price per day = $8

y = 40 + 8x

let

x = number of additional days

50 + 6x = 40 + 8x

collect like terms

50 - 40 = 8x - 6x

10 = 2x

Divide both sides by 2

x = 10/2

x = 5 days

Therefore, the number of additional days she wants to rent the car for, the two specials are equivalent is 5 days.

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3 0

1 year ago

Help!!!!!(ToT) thnaks

likoan [24]

The answer is A, 62

Hope this helps :)

5 0

3 years ago

According to the Knot, 22% of couples meet online. Assume the sampling distribution of p follows a normal distribution and answe

Ann [662]

Using the <em>normal distribution and the central limit theorem</em>, we have that:

a) The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.

b) There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.

c) There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.

<h3>Normal Probability Distribution</h3>

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.

By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1 - p)}{n}}$ , as long as $np \geq 10$ and $n(1 - p) \geq 10$ .

In this problem:

22% of couples meet online, hence p = 0.22.

A sample of 150 couples is taken, hence n = 150.

Item a:

The mean and the standard error are given by:

$\mu = p = 0.22$

$s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.22(0.78)}{150}} = 0.0338$

The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.

Item b:

The probability is <u>one subtracted by the p-value of Z when X = 0.25</u>, hence:

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

By the Central Limit Theorem:

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

$Z = \frac{0.25 - 0.22}{0.0338}$

Z = 0.89

Z = 0.89 has a p-value of 0.8133.

1 - 0.8133 = 0.1867.

There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.

Item c:

The probability is the <u>p-value of Z when X = 0.2 subtracted by the p-value of Z when X = 0.15</u>, hence:

X = 0.2:

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

$Z = \frac{0.2 - 0.22}{0.0338}$

Z = -0.59

Z = -0.59 has a p-value of 0.2776.

X = 0.15:

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

$Z = \frac{0.15 - 0.22}{0.0338}$

Z = -2.07

Z = -2.07 has a p-value of 0.0192.

0.2776 - 0.0192 = 0.2584.

There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.

To learn more about the <em>normal distribution and the central limit theorem</em>, you can check brainly.com/question/24663213

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2 years ago

A cone and a pyramid are both 3 inches in height. If the radius of the base of the cone is 4 in and the a side of the square bas

kotegsom [21]

I got the answer of the cone

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3 years ago

Read 2 more answers

A student repeatedly measures the mass of an object using a mechanical balance and gets the following values: 560 g, 562 g, 556

MrRissso [65]

Answer: 2.76 g

Step-by-step explanation:

The formula to find the standard deviation:-

$\sigma=\sqrt{\dfrac{\sum(x_i-\overline{x})^2}{n}}$

The given data values : 560 g, 562 g, 556 g, 558 g, 560 g, 556 g, 559 g, 561 g, 565 g, 563 g.

Then, $\overline{x}=\dfrac{\sum_{i=1}^{10} x_i}{n}\\\\\Rightarrow\ \overline{x}=\dfrac{560+562+556+558+560+556+559+561+565+563}{10}\\\\\Rightarrow\ \overline{x}=\dfrac{5600}{10}=560$

Now, $\sum_{i=1}^{10}(x_i-\overline{x})^2=0^2+2^2+(-4)^2+(-2)^2+0^2+(-4)^2+(-1)^2+1^2+5^2+3^2\\\\\Rightarrow\ \sum_{i=1}^{10}(x_i-\overline{x})^2=76$

Then, $\sigma=\sqrt{\dfrac{76}{10}}=\sqrt{7.6}=2.76$

Hence, the standard deviation of his measurements = 2.76 g

6 0

3 years ago