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ira [324]
4 years ago
12

Why is the product (-6)(-3) is positive?​

Mathematics
2 answers:
Mashutka [201]4 years ago
5 0
Because two negs make a positive
katen-ka-za [31]4 years ago
4 0
When you multiply two negatives you receive a positive.
When you subtract two negatives you get a positive.
Lastly, when you add two negatives you receive a negative.
Hope this helped :)
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A dartboard has 20 equally divided wedges and you are awarded the number of points in the section of your dark lands in if you a
Leto [7]

Answer:

Pr =0.75

Step-by-step explanation:

Given

See attachment for wedge

Required

P(x > 5)

The sample space of the wedge is:

S = \{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20\}

n(S) = 20

The outcomes greater than 5 are:

x > 5 = \{6,7,8,9,10,11,12,13,14,15,16,17,18,19,20\}

n(x>5) =15

So, the probability is:

Pr =\frac{(n(x>5)}{n(S)}

This gives:

Pr =\frac{15}{20}

Pr =0.75

6 0
3 years ago
ASAP Please thank you!
Pavel [41]

A cycle in the graph below is HDGH.

A cycle is where a vertices, or corner, can be traced back to itself. HDGH starts with H and traces back to H on the same line.

6 0
4 years ago
Tom works part-time at a theme park. last week he worked the following hours but he needs to complete his timesheet in decimals.
svetoff [14.1K]
He earns 6.92 per hour
8 0
3 years ago
Read 2 more answers
In a shipment of 56 vials, only 13 do not have hairline cracks. If you randomly select 3 vials from the shipment, in how many wa
Elden [556K]

Answer: 27434

Step-by-step explanation:

Given : Total number of vials = 56

Number of vials that do not have hairline cracks = 13

Then, Number of vials that have hairline cracks =56-13=43

Since , order of selection is not mattering here , so we combinations to find the number of ways.

The number of combinations of m thing r things at a time is given by :-

^nC_r=\dfrac{n!}{r!(n-r)!}

Now, the number of ways to select at least one out of 3 vials have a hairline crack will be :-

^{13}C_2\cdot ^{43}C_{1}+^{13}C_{1}\cdot ^{43}C_{2}+^{13}C_0\cdot ^{43}C_{3}\\\\=\dfrac{13!}{2!(13-2)!}\cdot\dfrac{43!}{1!(42)!}+\dfrac{13!}{1!(12)!}\cdot\dfrac{43!}{2!(41)!}+\dfrac{13!}{0!(13)!}\cdot\dfrac{43!}{3!(40)!}\\\\=\dfrac{13\times12\times11!}{2\times11!}\cdot (43)+(13)\cdot\dfrac{43\times42\times41!}{2\times41!}+(1)\dfrac{43\times42\times41\times40!}{6\times40!}\\\\=3354+11739+12341=27434

Hence, the required number of ways =27434

5 0
3 years ago
When x=4 what is the value of 8x
Naddik [55]
8*4=32 :) I hope this helps!
5 0
3 years ago
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