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3 years ago

HELP!! What is the value of x??

Mathematics

1 answer:

nordsb [41]3 years ago

6 0

Answer:

X = 41°

Step-by-step explanation:

<h3>Sum of interior Angles of a ∆ is 180°</h3>

$\angle1 + \angle2 + \angle3 = 180 \degree$

$Here \\ \angle1 = 105 \degree \\ \angle2 = ? \\ \angle3 = x$

$\angle4 = 146 \degree$

$here \: \angle2 + \angle4 = 180 \degree \\ \: (liner \: pair \: angle) degree\\ \angle2 + 146\degree = 180 \degree \\ \angle2 = 180 \degree - 146\degree \\ \angle2 = 34\degree$

$\angle1 + \angle2 + \angle3 = 180 \degree \\ 105\degree + 34\degree + x = 180\degree \\ 139\degree + x = 180\degree \\ x = 180\degree - 149\degree \\ x = 41\degree$

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Katarina [22]

Answer:

Equation: 65+1.50m=78.50

Solution to the equation= 9 minutes

Step-by-step explanation:

The parents have to give 65 dollars for their children to be under care till 6 o'clock. AFter 6 o'clock, they had to pay 1.50 dollars for one minute. This parent gives a total of 78.50 dollar for that day. So, the equation is :

65+1.50m=78.50

(btw, m stands for minutes)

So to solve the equation:

1.50m=13.50

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2 years ago

What expressions are equivalent to -56z + 28

makvit [3.9K]

Answer:

3b+b and 2(2b)

Step-by-step explanation:

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3 years ago

How would you solve for the question below???<br> step by step please and thanks.

Airida [17]

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-2 (2x + 3y) + (3x + 6y) = -2 (17) + 30

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2 years ago

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KonstantinChe [14]

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3 years ago

A major cab company in Chicago has computed its mean fare from O'Hare Airport to the Drake Hotel to be $28.75 , with a standard

bulgar [2K]

Answer:

Following are the solution to the given choices:

Step-by-step explanation:

Using chebyshev's theorem :

$\to P(|(x-\mu)|\leq k \sigma)\geq 1- \frac{1}{k^2}\\\\here \\ \to \mu=28.75\\\\ \to \sigma=4.44$

In point a)

$\to |(x-\mu)|= |20.17-28.75|=8.58 and \sigma=4.44 \\\\so, \\k= \frac{ |(x-\mu)| }{\sigma}$

$= \frac{8.58}{4.44} \\\\ =1.9 \\\\ =2$

$value= (1- \frac{1}{k^2}) \times 100 \% =75\%$

In point b)

$\to (1- \frac{1}{k^2}) \times 100 \% =84\% \\\\\to (1- \frac{1}{k^2})=0.84 \\\\\to \frac{1}{k^2} =0.16 \\\\\to \frac{1}{k}=0.4\\\\ \to k=2.5 \\\\\to |(x-\mu)| \leq k \sigma \\\\= |(x-\mu)|\leq 2.5 \times 4.44 \\\\ = |(x-\mu)|\leq 1.11 \\\\ = (28.75\pm 11.1) \\\\\to \text{fares lies between}(17.65,39.85)$

In point c)

$\to 99.7 \% \\ lie \ between=28.75 \pm z(.03)\times \sigma \\\\ =28.75\pm 2.97 \times 4.44\\\\=(28.75\pm 13.1)\\\\=(15.65,41.85)\\$

In point d)

using standard normal variate

$x=20.17\\\\ z=-2 \\\\x=37.13\\\\ z=2\\\\\to P(20.17$

8 0

3 years ago