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Goshia [24]
3 years ago
13

PLEASE HELP FINDING y IN TRIANGLE AND ROUND TO NEAREST TENTH TRIG!! HELP ASAP APLEASE

Mathematics
1 answer:
polet [3.4K]3 years ago
5 0

Answer:

<em>y = 12.3 cm</em>

Step-by-step explanation:

This is a right triangle, so we can use sin,cos, or tan to solve this.

First, with respect to the angle 35 degrees, we can say "y" is the "adjacent side".

Also, the hypotenuse is given as 15.

<em>Which trig ratio relates adjacent and hypotenuse? That is COS. So we can write and solve the equation below:</em>

<em>Cos(35)=\frac{y}{15}\\y=Cos(35)*15\\y=12.29</em>

<em />

<em>Rounded to 1 decimal place (nearest tenth) , </em>

<em>y = 12.3 cm</em>

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The quotient of k and 9 is greater than 1/3
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K/9 > 1/3
i believe this is how it is represented.
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3. Multiply the following binomials:<br> (2 Points)<br> (x - 3) (6x - 2) What’s is the answer?
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Answer:

3x^2-10x+3

Step-by-step explanation:

1. multiply each number

2. simplify

7 0
3 years ago
Find the slope of a line passing through the given points. 2,5) and (4,-7) help me
LekaFEV [45]

<em>-7 -5 / 4 -2 / -12/2 </em>

<em>-12/2 = -6 </em>

<em>Slope: -6x </em>

<em>5 = 2(-6) + b</em>

<em>5 = -12 + b </em>

<em>-12 + 17 = 5 </em>

<em>Slope intercept Form: y = -6x + 17</em>

7 0
3 years ago
A local little league has a total of 90 players of whom 20% are left-handed. How many left-handed players are there? Please help
german
18

First you want to change the 20% into a decimal which is .20
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7 0
3 years ago
A veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses
Licemer1 [7]

Answer:

Probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.

Step-by-step explanation:

We are given that a veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses with colic is 12 years. The average age of all horses seen at the veterinary clinic was determined to be 10 years. The researcher also determined that the standard deviation of all horses coming to the veterinary clinic is 8 years.

So, firstly according to Central limit theorem the z score probability distribution for sample means is given by;

                    Z = \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \bar X = average age of the random sample of horses with colic = 12 yrs

            \mu = average age of all horses seen at the veterinary clinic = 10 yrs

   \sigma = standard deviation of all horses coming to the veterinary clinic = 8 yrs

         n = sample of horses = 60

So, probability that a sample mean is 12 or larger for a sample from the horse population is given by = P(\bar X \geq 12)

   P(\bar X \geq 12) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } \geq \frac{12-10}{\frac{8}{\sqrt{60} } } ) = P(Z \geq 1.94) = 1 - P(Z < 1.94)

                                                 = 1 - 0.97381 = 0.0262

Therefore, probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.

4 0
3 years ago
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