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Find the value of in the triangle shown below 9, 7, and x around the triangle

enot [183]

Answer:

5.66 units

Step-by-step explanation:

A's per our discussion, x is a leg of right triangle.

Therefore, by Pythagoras theorem:

$x = \sqrt{ {9}^{2} - {7}^{2} } \\ = \sqrt{81 - 49} \\ = \sqrt{32}$

$x = \sqrt{ {16}^{2} - {12}^{2} } \\ = \sqrt{256 - 144} \\ = \sqrt{112}$

4 0

3 years ago

Ryan is trying a low-carbohydrate diet. He would like to keep the amount of carbs consumed in grams between the levels shown in

nydimaria [60]

Answer:

$50$

Ryan would like to eat more than 50 carbs per day, but no more than 150 carbs per day.

So, Ryan's total carb intake must be between 50 and 150 carbs.

Step-by-step explanation:

So he wants to keep his consumption of carbs between the inequalities:

$110$

So, let's solve both inequalities.

1)

$110$

Subtract 10 from both sides:

$100$

Divide both sides by 2:

$50$

2)

$2x+10$

Subtract 10 from both sides:

$2x$

Divide both sides by 2:

$x$

So, our inequality is now:

$110$

Since we solved the equations:

$50$

Written as a compound inequality, this is:

$50$

In other words, Ryan would like to eat more than 50 carbs per day, but no more than 150 carbs per day.

So, Ryan's total carb intake must be between 50 and 150 carbs.

4 0

3 years ago

Read 2 more answers

Max got another paycheck for $476.15 on April 20th.

Rufina [12.5K]

Answer:

$182.60

Step-by-step explanation:

4 0

3 years ago

Suiting at 6 a.m., cars, buses, and motorcycles arrive at a highway loll booth according to independent Poisson processes. Cars

dem82 [27]

Answer:

Step-by-step explanation:

From the information given:

the rate of the cars = $\dfrac{1}{5} \ car / min = 0.2 \ car /min$

the rate of the buses = $\dfrac{1}{10} \ bus / min = 0.1 \ bus /min$

the rate of motorcycle = $\dfrac{1}{30} \ motorcycle / min = 0.0333 \ motorcycle /min$

The probability of any event at a given time t can be expressed as:

$P(event \ (x) \ in \ time \ (t)\ min) = \dfrac{e^{-rate \times t}\times (rate \times t)^x}{x!}$

∴

(a)

$P(2 \ car \ in \ 20 \ min) = \dfrac{e^{-0.20\times 20}\times (0.2 \times 20)^2}{2!}$

$P(2 \ car \ in \ 20 \ min) =0.1465$

$P ( 1 \ motorcycle \ in \ 20 \ min) = \dfrac{e^{-0.0333\times 20}\times (0.0333 \times 20)^1}{1!}$

$P ( 1 \ motorcycle \ in \ 20 \ min) = 0.3422$

$P ( 0 \ buses \ in \ 20 \ min) = \dfrac{e^{-0.1\times 20}\times (0.1 \times 20)^0}{0!}$

$P ( 0 \ buses \ in \ 20 \ min) = 0.1353$

Thus;

P(exactly 2 cars, 1 motorcycle in 20 minutes) = 0.1465 × 0.3422 × 0.1353

P(exactly 2 cars, 1 motorcycle in 20 minutes) = 0.0068

(b)

the rate of the total vehicles = 0.2 + 0.1 + 0.0333 = 0.3333

the rate of vehicles with exact change = rate of total vehicles × P(exact change)

$= 0.3333 \times \dfrac{1}{4}$

= 0.0833

∴

$P(zero \ exact \ change \ in \ 10 minutes) = \dfrac{e^{-0.0833\times 10}\times (0.0833 \times 10)^0}{0!}$

P(zero exact change in 10 minutes) = 0.4347

c)

The probability of the 7th motorcycle after the arrival of the third motorcycle is:

$P( 4 \ motorcyles \ in \ 45 \ minutes) =\dfrac{e^{-0.0333\times 45}\times (0.0333 \times 45)^4}{4!}$

$P( 4 \ motorcyles \ in \ 45 \ minutes) =0.0469$

Thus; the probability of the 7th motorcycle after the arrival of the third one is = 0.0469

d)

P(at least one other vehicle arrives between 3rd and 4th car arrival)

= 1 - P(no other vehicle arrives between 3rd and 4th car arrival)

The 3rd car arrives at 15 minutes

The 4th car arrives at 20 minutes

The interval between the two = 5 minutes

For Bus:

P(no other vehicle other vehicle arrives within 5 minutes is)

= $\dfrac{6}{12} = 0.5$

For motorcycle:

$= \dfrac{2 }{12} = \dfrac{1 }{6}$

∴

The required probability = $1 - \Bigg ( \dfrac{e^{-0.5 \times 0.5^0}}{0!} \times \dfrac{e^{-1/6}\times (1/6)^0}{0!} \Bigg)$

= 1- 0.5134

= 0.4866

6 0

3 years ago

7а — За +2а — а = 16

dezoksy [38]

Answer:

(7a-3a)+(2a-a)=16

4a+a or 1a =16

5a=16

5a/5=16/5

a =16/5 or a=3.2

Step-by-step explanation:

solving like terms or substract are same

you only have to find value of uknown value which was "a" question

8 0

3 years ago