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Varvara68 [4.7K]
4 years ago
11

I need help figuring out what x is​

Mathematics
1 answer:
amm18124 years ago
8 0
There you go. Let me know if you have questions

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Solve each right triangle for ABC. <br> (Will award brainliest)
ladessa [460]

Answer:

6.57343137182

Step-by-step explanation:

The triangle is a right triangle

1. Use Pythagorean theorem

5.5² + 3.6² = c²

2. Solve

√ 43.21

4 0
3 years ago
Which of the following is not equal to cos⁡(520°)?
Gnoma [55]
The cosine repeats itself every 360 degrees, and it mirrors at 0 and 180 degrees. It inverts around 90 and 270.
So without using a calculator, you can tell that:

cos(520)=cos(520-360)=cos(160)
cos(160) = cos(180-20) = cos(180+20) = cos(200)
cos(160) is NOT equal to cos(20), it would be -cos(20).
cos(160) = -cos(20) = -cos(-20)

So C is the only one unequal.
8 0
3 years ago
Read 2 more answers
There are 534 car parking spaces but 478 spaces are occupied,round to the nearest 10
suter [353]

Answer:

60

Step-by-step explanation:

If 478 out of 534 parking spaces are taken, then you will be <u>subtracting</u>. When you subtract 534 and 478 (534-478), then you get 54. Remember 5 and up round to 10. 4 and below round to 0. So 56 rounds to 60 because 6 is greater than 5.

Hope it helps!

8 0
4 years ago
Josey got an answer of 167r4 for 505 divided by 3.Explain and correct his error.
algol [13]

    167r1
3/ 505
    3
--------
    20
    18
____
    25
    24
_____
       1 remainder
    167r1
3/ 505
    3
--------
    20
    18
____
    25
    24
_____
       1 remainder
3 0
4 years ago
Read 2 more answers
A standard number cube is rolled 288 times. Predict how many times a 3 or a 5 will be the result.
Anvisha [2.4K]
The events of rolling a 3 or a 5 are disjoint, so \mathbb P\bigg((X=3)\cup(X=5)\bigg)=\mathbb P(X=3)+\mathbb P(X=5).

Each face has a \dfrac16 probability of occurring, so the event of interest occurs with probability \dfrac16+\dfrac16=\dfrac13. The number of times a 3 or 5 is rolled across 288 trials has a binomial distribution, \mathcal B\left(288,\dfrac13\right).

Recall that the expected value of a binomial distribution \mathcal B(n,p) is np. So in this experiment, we should expect to witness approximately \dfrac{288}3=96 instances of rolling a 3 or a 5.
5 0
3 years ago
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