Is 1/2 greater than 2/5

A football stadium sells regular and box seating. There are twelve times as many

Vera_Pavlovna [14]

Answer:

There are 801 box seats and 9612 regular seats.

explanation:

Let the number of box seats be x,

Then the regular seats is 12x

The sum of seats equals to 10,413

<u>Solve:</u>

12x + x ➙ 10,413

13x ➙ 10413

x ➙ 801

There are 801 box seats.

<u>Find for regular seats:</u>

➙ 12(x)

➙ 12(801)

➙ 9612

There are 9612 regular seats.

6 0

1 year ago

Read 2 more answers

If f(x) = 3x + 10 and g(x) = 2x - 4, find (f - g)(x).

torisob [31]

<h3>Answer: Choice A) x+14</h3>

=======================================

Work Shown:

(f-g)(x) = f(x) - g(x)

(f-g)(x) = (f(x)) - (g(x))

(f-g)(x) = (3x+10) - (2x-4)

(f-g)(x) = 3x+10 - 2x+4

(f-g)(x) = (3x-2x) + (10+4)

(f-g)(x) = x+14

4 0

3 years ago

An ice cream factory makes 100 quarts of ice cream in 5 hours how many quarts could be made in 36 hours

I am Lyosha [343]

Answer:

720 Quarts

Step-by-step explanation:

$\frac{100}{5} = \frac{x}{36}$

cross multiply

$\frac{100}{5}*\frac{36}{36} = \frac{x}{36}*\frac{5}{5}\\\\\frac{3,600}{180}=\frac{5x}{180}$

now that they are both divided by 180, we can eliminate it

$3,600=5x\\\\\frac{3,600}{5}=\frac{5x}{5}\\\\720=x$

7 0

2 years ago

Read 2 more answers

Hospital records show that a certain surgical procedure takes on an average 111.6 minutes with a standard deviation of 2.8 minut

Arisa [49]

Answer:

At least 37.74% of the procedures takes between 97.6 and 125.6 minutes

Step-by-step explanation:

The percentage is the area of the standard normal distribution curve between the values 97.6 and 125.6

The standard normal variate of the given data is found as

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

Thus for the given values Z is calculated as under

$Z_1=\frac{97.6-111.6}{2.8}=-5$

Similarly for the other value we have

$Z_2=\frac{125.6-111.6}{2.8}=5$

Thus the area between the calculated values can be found from standard normal distribution table to be 37.74%

8 0

3 years ago

Find the logarithmic function y = logbx that passes through the points

katovenus [111]

$b\in(0,\ 1)\ \cup\ (1,\ \infty)\\x > 0\\y\in\mathbb{R}\\--------------------------\\\\y=\log_bx\\\\For\ (1,\ 0)\to x=1,\ y=0.\ Substitute:\\\\\log_b1=0\to b^0=1\to b\in(0,\ 1)\ \cup\ (1,\ \infty)\\\\For\ (116,\ 2)\to x=116,\ y=2.\ Substitute:\\\\\log_b116=2\to b^2=116\to b=\sqrt{116}\\\to b=\sqrt{4\cdot29}\to b=\sqrt4\cdot\sqrt{29}\to b=2\sqrt{29}\\\\For\ (4,\ -1)\to x=4,\ y=-1.\ Substitute:\\\\\log_b4=-1\to b^{-1}=4\to b=\dfrac{1}{4}$

Different values of b.

<h3>Answer: There is no logarithmic function whose graph goes through given points.</h3><h3 />

Maybe the second point is $\left(\dfrac{1}{16},\ 2\right)$

Substitute:

$\log_b\dfrac{1}{16}=2\to b^2=\dfrac{1}{16}\to b=\sqrt{\dfrac{1}{16}}\to b=\dfrac{1}{4}$

<h3>Then we have the answer:</h3>

$\boxed{y=\log_{\frac{1}{4}}x}$

6 0

3 years ago