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zysi [14]
3 years ago
7

Please help me i’ll give brainliest

Mathematics
1 answer:
liraira [26]3 years ago
6 0

Answer:

105 degrees

Step-by-step explanation:

You need to understand the properties of corresponding angles and supplementary angles.

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How many seconds equals 4.2 hours?<br><br> Enter your answer in the box.<br><br> ___s
sergiy2304 [10]
4.2× 3600 = 15120 seconds
8 0
3 years ago
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A businessman was comparing his company's profits for two consecutive months. In November, n, were 0.04 thousand dollars more th
Firdavs [7]
N and d are both numbers of thousands of dollars.
Thus, if n = 1, that means $1000.

Here n = d + 0.4.
Note that in C, n+0.04=n is completely wrong.
Similarly, in D, d = 0.04 = n is completely wrong.
The "combined profit amount" for Nov. and Dec. is n + d = 3.15.

Only A matches this info.  Your answer is A.
5 0
3 years ago
Gina has 5 2/6 feet of silver ribbon and 2 4/6 of gold ribbon
topjm [15]

Answer:

5 2/6 + 2 4/6 = 8

Step-by-step explanation:

So Gina has 8 feet of ribbon in total.

I hope this is what you were asking for!! :)

Have a good day!

Brainliest PLEASE!!?!?!?!??!?!!?!!?!?!?

8 0
3 years ago
I am just completely lost and am unsure of how to approach this problem.
Anastasy [175]

Answer:

68°

Step-by-step explanation:

As BA⊥BD, so ∠ABD = 90°

also, ∠ABD = ∠CBD + ∠ABC

⇒ 90° = 4x + 52° + 8x - 10°

⇒  12x = 48°

⇒  x = 4°

now,

∠CBD = 4x + 52°

          = 4(4°) + 52°

          = 68°

5 0
3 years ago
Consider a large shipment of 400 refrigerators, of which 40 have defective compressors. If X is the number among 15 randomly sel
Vlad [161]

Answer:

The required probability is 0.94

Step-by-step explanation:

Consider the provided information.

There are 400 refrigerators, of which 40 have defective compressors.

Therefore N = 400 and X = 40

The probability of defective compressors is:

\frac{40}{400}=0.10

It is given that If X is the number among 15 randomly selected refrigerators that have defective compressors,

That means n=15

Apply the probability density function.

P(X=x)=^nC_xp^x(1-p)^{n-x}

We need to find P(X ≤ 3)

P(X\leq3) =P(X=0)+P(X=1)+P(X=2)+P(X=3)\\P(X\leq3) =\frac{15!}{15!}(0.1)^0(1-0.1)^{15}+\frac{15!}{14!}(0.1)^1(1-0.1)^{14}+\frac{15!}{13!2!}(0.1)^2(1-0.1)^{13}+\frac{15!}{12!3!}(0.1)^3(1-0.1)^{12}\\

P(X\leq3) =0.944444369992\approx 0.94

Hence, the required probability is 0.94

4 0
3 years ago
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