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vodomira [7]
3 years ago
8

Question

Mathematics
2 answers:
steposvetlana [31]3 years ago
7 0

Answer:

45°, 75°, 61°, 44°, 22°, 86°

Step-by-step explanation:

They have to be angles less than 90°.

babunello [35]3 years ago
4 0

Answer:

  (c)  75°, 61°, 44°

  (d)  22°, 86°, 72

Step-by-step explanation:

The largest angle must be less than 90°, and the sum of angles must be 180°. These conditions are met by the lists of angles shown above.

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yulyashka [42]

Answer:

5

Step-by-step explanation:

5 (l)x4(h)=20

(l)=laps

(h)=Hours

3 0
4 years ago
Read 2 more answers
A golf ball is hit on the moon so that
storchak [24]

I am just not sure if I did this properly. My professor hasn't really gone over when to use the range equation but I would assume range would equal the distance traveled therefore can be used for this problem. If not the how would I go about solving this? I did 1/6*9.8=1.63 for g on the moon Then used R=(Vi^2sin2theta)/g so on the moon R=((28^2)sin2*35)/1.63=452m

Please Brainliest..? thx

5 0
3 years ago
C. If 48 increases by 130%
alexandr402 [8]

Answer:

69%

Step-by-step explanation:

5 0
3 years ago
What is the value in cents of 10 quarters and 2 dimes
BaLLatris [955]

Answer:

250+20 = <u><em>270 cents or $2.70</em></u>

Step-by-step explanation:

Each quarter is 25 cents

Each dime is 10 cents

<h2><em><u>GIVEME BRAINLIEST</u></em></h2>

5 0
3 years ago
A chip used in the production of particleboard (Question 4) has a 15% chance of containing excessive bark. What is the probabili
salantis [7]

Answer:

0.2

Step-by-step explanation:

Given that:

P(particleboard with excessive bark) = 0.15

The P(that particleboard do not have excessive bark) = 1 - 0.15

i.e the population proportion number of success p = 0.85

The sample size (n) is given to be = 1000

To find the P(X ≥ 860)

Since the sample size is large, we will apply the normal approximation of binomial distribution to treat this question.

The population mean \mu = n \times p

\mu = 1000 \times 0.85

\mu =850

The population standard deviation \sigma = \sqrt{n*p(1-p)

\sigma = \sqrt{1000*0.85*(1-0.85)}

\sigma = \sqrt{1000*0.85*(0.15)}

\sigma = \sqrt{127.5}

\sigma = 11.2916

Let X be the random variable which obeys a normal distribution;

Then;

P(X \ge 860) = P(Z\ge \dfrac{x- \mu}{\sigma})

P(X \ge 860) = P(Z\ge \dfrac{860- 850}{11.2916})

P(X \ge 860) = P(Z\ge \dfrac{10}{11.2916})

P(X ≥ 860) = P(Z ≥ 0.8856)

P(X ≥ 860) = 1 - P(Z ≤ 0.8856)

From  z table

P(X ≥ 860) = 1 - 0.8122

P(X ≥ 860) = 0.1878

P(X ≥ 860) \simeq 0.2

Thus, the probability of having more than 860 bark-free chips in a batch of 1,000 = 0.2

6 0
3 years ago
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