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Flura [38]
2 years ago
12

How many tickets were sold after June 15th?

Mathematics
2 answers:
solong [7]2 years ago
6 0
A) about 45 dhhdhdhduejebejenebejekwnee
MatroZZZ [7]2 years ago
3 0
A) about 45

If we sold around 90, it would NOT be shown in the graph. And starting on June 15, we sold (per say 50). On June 19 we end with approximately 100. So 45 would be the most probable answer.

Hope this helps!
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Prove that:<br><br>cos20°cos40°cos80°=1/8​
Naya [18.7K]

Answer:

see explanation

Step-by-step explanation:

Using the double angle identity for sine

sin2x = 2sinxcosx

Consider left side

cos20°cos40°cos80°

= \frac{1}{2sin20} (2sin20°cos20°)cos40°cos80°

= \frac{1}{4sin20} (2sin40°cos40°)cos80°

= \frac{1}{4sin20} (sin80°cos80° )

= \frac{1}{8sin20} (2sin80°cos80° )

= \frac{1}{8sin20} . sin160°

= \frac{1}{8sin20} . sin(180 - 20)°

= \frac{1}{8sin20} . sin20°

= \frac{1}{8} = right side , thus proven

7 0
2 years ago
Read 2 more answers
It costs 4 tokens to park in a garage for an hour. how many hours can you park with 30 tokens?
babunello [35]

Read the question carefully: it costs 4 tokens to park in a garage for an hour.

We will apply the unitary method to solve this question

It costs 4 tokens to park in a garage for 1 hour

Find how many hours can park in a garage for 1 token

If it costs 4 token to park in a garage for 1 hour

Then it will cost 1 token to park in a garage for 1/4 hour

Step2:

With 20 token we can park in a garage for (1/4) * 20

= 5 hours

So, we can park for 5 hours with 20 tokens.

Another method

If we take twenty tokens and divide them into groups of four, we will find that we are left with five groups of tokens. Each group of tokens represents an hour of parking time. This will give us five groups, or five hours, total.

So, we can park for 5 hours with 20 tokens
5 0
3 years ago
Read 2 more answers
Help help help help help
nekit [7.7K]

Answer:

C, C is the answer

5 0
3 years ago
The combined math and verbal scores for students taking a national standardized examination for college admission, is normally d
kipiarov [429]

Answer:

The minimum score that such a student can obtain and still qualify for admission at the college = 660.1

Step-by-step explanation:

This is a normal distribution problem, for the combined math and verbal scores for students taking a national standardized examination for college admission, the

Mean = μ = 560

Standard deviation = σ = 260

A college requires a student to be in the top 35 % of students taking this test, what is the minimum score that such a student can obtain and still qualify for admission at the college?

Let the minimum score that such a student can obtain and still qualify for admission at the college be x' and its z-score be z'.

P(x > x') = P(z > z') = 35% = 0.35

P(z > z') = 1 - P(z ≤ z') = 0.35

P(z ≤ z') = 1 - 0.35 = 0.65

Using the normal distribution table,

z' = 0.385

we then convert this z-score back to a combined math and verbal scores.

The z-score for any value is the value minus the mean then divided by the standard deviation.

z' = (x' - μ)/σ

0.385 = (x' - 560)/260

x' = (0.385×260) + 560 = 660.1

Hope this Helps!!!

8 0
3 years ago
Find the surface area of the cylinder represented by the trunk of the tree.
Montano1993 [528]

Answer:

43.98

Step-by-step explanation:

8 0
2 years ago
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