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Elenna [48]
3 years ago
15

What is the measure of ABC?

Mathematics
1 answer:
NeX [460]3 years ago
7 0

A. 44

Step-by-step explanation:

An inscribed angle measures the middle of this arc.

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Triangle OBC is a right triangle drawn on a coordinate plane. Side OC lies on the x-axis, the coordinates of point O are (0, 0),
Rudik [331]
Long leg = 8 → opposite
short leg = 6 → adjacent
hypotenuse = ?

8² + 6² = c²
64 + 36 = c²
100 = c²
√100 = √c²
10 = c

sin ∠BOC = opposite / hypotenuse
sin ∠BOC = 8 / 10 
sin ∠BOC = 0.80

tan ∠BOC = opposite / adjacent
tan ∠BOC = 8 / 6
tan ∠BOC = 1.33

3 0
2 years ago
The director of marketing at a large company wants to determine if the amount of money spent on internet marketing is a good pre
san4es73 [151]
Given:
Profit = 372.60 + 17.2(advertising dollars)
Advertising dollars = 1,020
profit = 17,500

17,500 + residual value = 372.60 + 17.2(1,020)
17,500 + residual value = 372.60 + 17,544
17,500 + residual value = 17,916.60
residual value = 17,916.60 - 17,500
residual value = 416.60

residual value is 416.60 or 417 rounded off to the nearest integer
3 0
3 years ago
What is the equivalent of 6x-8y?​
liberstina [14]
Plug y=-5 into the equation
6x+8(-5)=-22
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6 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
Consider the given density curve.
Gemiola [76]

Answer:

11 is best answer for it ....,..........

4 0
2 years ago
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