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xxTIMURxx [149]
3 years ago
5

Lines CD and DE are tangent to circle A shown below: Lines CD and DE are tangent to circle A and intersect at point D. Arc CE me

asures 112 degrees. Point B lies on circle A. If Arc CE is 112°, what is the measure of ∠CDE? (4 points) Question 5 options: 1) 124° 2) 136° 3) 68° 4) 56°
Mathematics
1 answer:
Verdich [7]3 years ago
7 0

Answer:

Option (3)

Step-by-step explanation:

By the theorem, angle between two tangents intersecting each other outside the circle is half of the difference between intercepted arcs.

From the figure attached,

Two tangents CD and DE are intersecting each other at point D outside the circle A.

Two intercepted arcs are minor arc CE and major arc CBE,

Measure of arc CE = 112°

Therefore, measure of major arc CBE = 360° - 112°

                                                               = 248°

m(∠CDE) = \frac{1}{2}(\text{arcCBE}-\text{arcCE})

                 = \frac{1}{2}(248-112)

                 = \frac{1}{2}\times 136

                 = 68°

Option (3) will be the correct option.

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Based on the data in the two-way table, the probability of being 25-35 years and having a hemoglobin level above 11 is __ a.29 b
Alina [70]

The probability of being 25-35 years and having a haemoglobin level above 11 is 34%.

The probability of having a haemoglobin level above 11 is 36%.

Being 25-35 years and having a hemoglobin level above 11 are not dependent on each other.

<h3>What are the probabilities?
</h3>

Probability determines the odds that a random event would occur. The odds of the event happening lie between 0 and 1.

The probability of being 25-35 years and having a haemoglobin level above 11 = number of people between 25 - 35 that have a level above 11 / total number of people between 25 - 35

44 / 128 = 34%

The probability of having a haemoglobin level above 11  = number of people with a level above 11 / total number of respondents

153 / 429 = 36%

To learn more about probability, please check: brainly.com/question/13234031

#SPJ1

4 0
2 years ago
You supervise the feeding of dogs at a kennel you use an average of 10 1/2 pounds of dog food each day.at the rate how many poun
liq [111]

Answer:

73.5 pounds of dog food in 7 days

Step-by-step explanation:

10 1/2 x 7

10x 7 =70

1/2 x 7 = 3.5

70 + 3.5 =73.5 pounds of dog food in 7 days

6 0
2 years ago
5(2n - 4) = 9(n + 2)​
Wewaii [24]
1. Distribute
5(2n-4)
10n-20

9(n+2)
9n+18

10n-20=9n+18

2. Combine like terms
combine the terms with “n”, and then combine the other numbers.

Hope this helps!
8 0
2 years ago
If you are driving 20 meters per second, what would be the equivalent of that in miles per hour?
melomori [17]

\frac{20 meters}{1 second}  \frac{60 seconds}{1 minute}  \frac{60 minutes}{1 hour}

the fractions are set up so that each unit cancels out until its only meters/hour

you can multiply all the numerators together to get 72000 and all the denominators to get 1

72000 meters/1 hour

72000 meters in 1 hour

hope this helped

8 0
2 years ago
WILL GIVE BRAINLIEST
stira [4]

The air force plane travelled for a total of 6.9 hours.

<u>SOLUTION: </u>

Given, An Air Force plane left Nairobi and flew west at an average speed of 159 mph.  A cargo plane left sometime later flying in the same direction at an average speed of 207 mph.  After flying for 5.3 hours the cargo plane caught up with the Air Force plane.  

We have to find the number of hours the Air Force plane flew before the cargo plane caught up.

Now, we know that, \text { distance travelled }=\text { speed } \times \text { time }

For air force plane \rightarrow \text { distance travelled }=159 \mathrm{x} \text { time taken by air force plane }

\text { For cargo plane } \rightarrow \text { distance travelled }=207 \times 5.3 \text { hours }

Now, when cargo plane caught the air force plane the distance travelled by the planes will be equal.

So, 159 \times \text { time taken by air force plane }=207 \times 5.3

\Rightarrow =\frac{207 \times 5.3}{159}

\Rightarrow =\frac{1097.1}{159}

Time taken by air force plane = 6.9  hrs

6 0
3 years ago
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