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Serjik [45]
2 years ago
15

Convert (2,210°) to rectangular form.

Mathematics
1 answer:
8090 [49]2 years ago
5 0

your answer would be (-√3, -1)

i hope it helped !!

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What is71 3/4 over 100 as a decimal<br><br><br> pls explain
lora16 [44]

Answer:

3 71/100 = 3.71

Step-by-step explanation:

To convert this mixed number (fraction) to a decimal just follow these two steps:

Step 1: divide numerator (71) by the denominator (100): 71 ÷ 100 = 0.71

Step 2: add this value to the the integer part: 3 + 0.71 = 3.71. So,

3 71/100 = 3.71

7 0
2 years ago
Read 2 more answers
Assume that 1200 births are randomly selected and exactly 595 of the births are girls. Use subjective judgment to determine whet
emmainna [20.7K]

Answer:

Lets assume that 1200 births are randomly selected and exactly 595 of the births are girls.

In percentage this is : \frac{595}{1200}\times100=49.58\%

This is approx 50%. We can say that almost random 50% are girls.

So, It is unlikely because the probability of this particular outcome is very​ small, considering all of the other possible outcomes.

Part B: It is not unusual because 595 is about the number of girls expected.

6 0
3 years ago
Please help me<br> it means a lot
fgiga [73]
1. 5
2. 5
3. 10
4. 25
5. -11

6.1/4
7. 1
8. -1

7 0
3 years ago
-.27(x-2.9t)^2+2.56 in standard form?<br> Please provide steps. Thank you!
kirill [66]

Answer:

-\frac{27(-\frac{20t}{10}+x)^{n}  }{50} +2.56

Step-by-step explanation:

6 0
1 year ago
Samples of rejuvenated mitochondria are mutated(defective) in 1% ofcases. Suppose 15 samples are studied, and they can be consid
Bezzdna [24]

Answer:

A)0.86

B)0.99

C)6.047E-13

Step-by-step explanation:

Since p = 0.01 and q= 0.99

Number of event n = 15

Using binomial expansion

A) Probability (No samples are mutated) = 15C0 (p)^0 * (q)^15

= 1*(0.01)^0*(0.99)^15 = 0.86

B)Probability (At most one sample is mutated) = 15C0 * p^0 *q^15 + 15C1 * p^1 * q^14 = 1*(0.01)^0*(0.99)^15 + 15C1 * 0.01 * 0.99^14 = 0.86 + 15*0.01*0.8687 = 0.86 + 0.13 = 0.99

C) Probability (More than half the samples are mutated) = 15C8* 0.01^8 * 0.99^7 + 15C9* 0.01^9 * 0.99^6 +15C10* 0.01^10 *0.99^5 + 15C11 * 0.01^11 * 0.99^4 + 15C12 * 0.01^12 * 0.99^3 + 15C13 * 0.01^13 * 0.99^ 2 + 15C14 * 0.01^14 * 0.99^1 + 15C15 * 0.01^15 * 0.99^0

Probability (More than half the samples are mutated) = 6E-13 + 4.71E-15 + 2.86E-17 + 1.31E-19 + 4.41E-22 + 1.03E-24 + 1.49E-27 + 1E-30 = 6.047E-13

3 0
3 years ago
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