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Katena32 [7]
3 years ago
8

Please help me with this answer

Mathematics
1 answer:
victus00 [196]3 years ago
6 0

Answer: I believe it is c 4

Step-by-step explanation:

but im not sure

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64×15=32×30 true or false
Dima020 [189]

\large \mathfrak{Solution : }

Yes, it's true because the value of LHS = RHS if we solve it .

  • 64 \times 15 = 32 \times 30

  • 90 = 90
4 0
3 years ago
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Is :<br> -.6 <br> IRRATIONAL?<br> yes<br> or <br> no
Anastaziya [24]

Answer:

I would say -6 is an irrational number.

Step-by-step explanation:

Either way, -6 is a rational number, because it can be expressed as a fraction where the numerator and denominator are integers and the denominator doesn't equal 0.

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3 years ago
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Is 7 a foctor of 93?<br>Explain your answer using words.​
Sveta_85 [38]

No, because 93/7 is not a rational number

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4 years ago
Normal Distribution. Cherry trees in a certain orchard have heights that are normally distributed with mu = 112 inches and sigma
Lubov Fominskaja [6]

Answer:

The probability that a randomly chosen tree is greater than 140 inches is 0.0228.

Step-by-step explanation:

Given : Cherry trees in a certain orchard have heights that are normally distributed with \mu = 112 inches and \sigma = 14 inches.

To find : What is the probability that a randomly chosen tree is greater than 140 inches?

Solution :

Mean - \mu = 112 inches

Standard deviation - \sigma = 14 inches

The z-score formula is given by, Z=\frac{x-\mu}{\sigma}

Now,

P(X>140)=P(\frac{x-\mu}{\sigma}>\frac{140-\mu}{\sigma})

P(X>140)=P(Z>\frac{140-112}{14})

P(X>140)=P(Z>\frac{28}{14})

P(X>140)=P(Z>2)

P(X>140)=1-P(Z

The Z-score value we get is from the Z-table,

P(X>140)=1-0.9772

P(X>140)=0.0228

Therefore, the probability that a randomly chosen tree is greater than 140 inches is 0.0228.

5 0
3 years ago
Can someone help? It’s about 2 column proofs
Gwar [14]

Step-by-step explanation:

1. AB = BC (B is the midpoint of AC)

2. DE = EF (E is the midpoint of DF)

3. EB is common

4. ∠ABE = ∠CBE; ∠BED = ∠BEF (EB⊥AC, EB⊥DF)

5. ΔDEB ≅ ΔFEB (RHS)

6. DB = FB (corresponding ∠s of ≅ Δs)

7. ∠EFB = ∠CBF; ∠EDB = ∠ABD (alternate interior angles, AC║DF)

8. ΔABD ≅ ΔCBF (SAS)

4 0
3 years ago
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