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Wittaler [7]
4 years ago
5

Balancing the equations

Mathematics
1 answer:
erastovalidia [21]4 years ago
8 0
The first one is 1, 1, 1. the second one is 3, 4, 1, 4.
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Find the measure of the missing angles.
Sergio039 [100]

Answer:

d = 105° , e = 32° , f = 43°

Step-by-step explanation:

d and 105° are vertically opposite angles and are congruent , then

d = 105°

e and 32° are vertically opposite angles and are congruent , then

e = 32°

105° , e and f lie on a straight line and sum to 180° , that is

105° + e + f = 180°

105° + 32° + f = 180°

137° + f = 180° ( subtract 137° from both sides )

f = 43°

8 0
2 years ago
Read 2 more answers
Gracie, Mary, and Nancy each have a small collection of seashells. Gracie has 5 more than 1 1/4 times the number of shells Mary
Zolol [24]
G = 1.25m + 5
n = 1.50n + 1
g = n

1.25m + 5 = 1.50m + 1 <== ur equation
5 - 1 = 1.50m - 1.25m
4 = 0.25m
4 / 0.25 = m
16 = m <== mary has 16 <==

g = 1.25m + 5....g = 1.25(16) + 5....g = 25 <==Gracie has 25 <==
n = 1.50m + 1...n = 1.50(16) + 1.....n = 25 <==Nancy has 25 <==
8 0
4 years ago
Find ratio of 378 to 945
Bingel [31]

Answer:

2:5

Step-by-step explanation:

We can simplify the ratio 378 : 945 by dividing both terms by the greatest common factor (GCF).

The GCF of 378 and 945 is 189.

Divide both terms by 189.

378 ÷ 189 = 2

945 ÷ 189 = 5

Therefore:

378 : 945 = 2 : 5

5 0
3 years ago
Read 2 more answers
For a binomial probability distribution, it is unusual for the number of successes to be less than μ − 2.5σ or greater than μ +
kirill [66]

Answer:The correct option is a)

The upper limit of successes that would be deemed to be usual is 5, so more than 5 successes would be unusual.

Step-by-step explanation:

The mean of a binomial distribution

μ= np where n=10 and p= 0.2 q= 1-p, q= 1-0.2=0.8

μ= 10×0.2=2

σ=√npq

σ=√10×0.2×0.8=1.26

Is unusual for the number of success to be greater than μ + 2.5σ.

= 2+2.5(1.26)

=5 approximately.

So it is unusual for it to be greater than 0.5. The right option Is a)

7 0
3 years ago
Three pumps can remove a total of 1700 gallons of water per minute from a flooded mineshaft. If engineers want to remove at leas
lesya [120]

In total, they would need 4 pumps. If we divide 55 by 17, we get just more than 3. This little more would need an entire pump. So the answer 4 total, 3 added

4 0
3 years ago
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