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velikii [3]
3 years ago
8

How old are all of y'all if I'm 10 and if you are 12 then what is 12times10

Mathematics
2 answers:
Serjik [45]3 years ago
3 0
12 times ten equals one hundred twenty
Verizon [17]3 years ago
3 0
The answer would be 120 because you have to put 12 on the top and 10 on the bottom
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Segment addition, how do u solve it? Step by step.
weeeeeb [17]

Answer:

x=7  DO=25  OG=35

Step-by-step explanation:

4x-3+2x+21=60            4(7)-3                  2(7)+21

4x+2x   -3+21                 28-3=25             14+21=35

6x+18=60                     DO=25                  OG=35

60-18=42

6x=42

42/6=7

x=7

5 0
3 years ago
I'LL GIVE BRAINLIST<br><br> Select all the rates that are unit rates.
Alex Ar [27]

Answer:

B and D

Step-by-step explanation:

5 0
3 years ago
3500ca = 1fp
MAXImum [283]

Answer: 1fp = 525 minutes

           35fp = 18,375 minutes

<u>Step-by-step explanation:</u>

Solve this using conversions

\dfrac{30\ minutes}{200\ ca } \times \dfrac{3500\ ca}{1\ fp}= 525\ minutes\ per\ fp

For 35 fp, multiply 35 fp by 525 minutes per 1 fp = 18,375

5 0
3 years ago
Helpppp!!
Marysya12 [62]
For the experimental probability you must record the data that you collect by flipping your own coin, then you must find the probability of landing on either side. For example, the theoretical probability for the coin toss it will be 50% chance for either side. For the experimental it depends on your own results.

3 0
3 years ago
(a) A lamp has two bulbs, each of a type with average lifetime 1400 hours. Assuming that we can model the probability of failure
Temka [501]

Answer:

For first lamp ; The resultant probability is 0.703

For both lamps; The resultant probability is 0.3614

Step-by-step explanation:

Let X be the lifetime hours of two bulbs

X∼exp(1/1400)

f(x)=1/1400e−1/1400x

P(X<x)=1−e−1/1400x

X∼exp⁡(1/1400)

f(x)=1/1400 e−1/1400x

P(X<x)=1−e−1/1400x

The probability that both of the lamp bulbs fail within 1700 hours is calculated below,

P(X≤1700)=1−e−1/1400×1700

=1−e−1.21=0.703

The resultant probability is 0.703

Let Y be a lifetime of another lamp two bulbs

Then the Z = X + Y will follow gamma distribution that is,

X+Y=Z∼gamma(2,1/1400)

2λZ∼

X+Y=Z∼gamma(2,1/1400)

2λZ∼χ2α2

The probability that both of the lamp bulbs fail within a total of 1700 hours is calculated below,

P(Z≤1700)=P(1/700Z≤1.67)=

P(χ24≤1.67)=0.3614

The resultant probability is 0.3614

8 0
3 years ago
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