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7 months ago

Find the values of x and y

Mathematics

1 answer:

ioda7 months ago

4 0

X= 20 & Y= 70 in right-angled triangle .

What is right angle triangle?

A right-angled triangle is a particular kind of triangle in which one of the angles is 90 degrees.
The combined angles of the other two are 90 degrees. Perpendicular and the triangle's base make up the sides that the right angle is formed from.
The longest of the three sides, known as the hypotenuse, is the third side.

2X + 140 = 180

2X = 40°

X = 20°

X + Y = 90°

Y = 70°

Learn more about right-angled triangle

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3s+2t−3 =c<br> −7s−5t =4<br>

irinina [24]

Answer:

−4

s−t+

−4

Step-by-step explanation:

Let's solve for c.

3s+2t−3c−7s−5t=4

Step 1: Add 4s to both sides.

−3c−4s−3t+4s=4+4s

−3c−3t=4s+4

Step 2: Add 3t to both sides.

−3c−3t+3t=4s+4+3t

−3c=4s+3t+4

Step 3: Divide both sides by -3.

−3c

−3

4s+3t+4

−3

−4

s−t+

−4

3 0

2 years ago

Ravi bought 50kg rice at the rate it tk40 per kg and sold it at the rate of tk44 per kg. What is the amount of profit or loss?

ololo11 [35]

Here is your Answer

Step-by-step explanation:

Ravi bought rice = 50kg

Rate it =40kg

Sold it at the rate = = 44kg

<h3>Answer: -34</h3><h2 /><h2 /><h2>Mark me as brainlist and give me 20 thx pls</h2>

6 0

2 years ago

1. A report from the Secretary of Health and Human Services stated that 70% of single-vehicle traffic fatalities that occur at n

Nuetrik [128]

Using the binomial distribution, it is found that there is a 0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

For each fatality, there are only two possible outcomes, either it involved an intoxicated driver, or it did not. The probability of a fatality involving an intoxicated driver is independent of any other fatality, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

70% of fatalities involve an intoxicated driver, hence $p = 0.7$ .

A sample of 15 fatalities is taken, hence $n = 15$ .

The probability is:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)$

Hence

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{15,10}.(0.7)^{10}.(0.3)^{5} = 0.2061$

$P(X = 11) = C_{15,11}.(0.7)^{11}.(0.3)^{4} = 0.2186$

$P(X = 12) = C_{15,12}.(0.7)^{12}.(0.3)^{3} = 0.1700$

$P(X = 13) = C_{15,13}.(0.7)^{13}.(0.3)^{2} = 0.0916$

$P(X = 14) = C_{15,14}.(0.7)^{14}.(0.3)^{1} = 0.0305$

$P(X = 15) = C_{15,15}.(0.7)^{15}.(0.3)^{0} = 0.0047$

Then:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.2061 + 0.2186 + 0.1700 + 0.0916 + 0.0305 + 0.0047 = 0.7215$