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luda_lava [24]
3 years ago
14

How we can find decimal between 2.15 and 2.17 ?

Mathematics
2 answers:
Bess [88]3 years ago
6 0
Find a number between 2.15 and 2.17 so one of the answers could be 2.16
Nadusha1986 [10]3 years ago
3 0
2.16-2.15=0.02

The difference is 0.02
To get the number between you divide the difference by 2 then add it to the lower number.
0.02/2=0.01

2.15+0.01=2.16

The number between 2.15 and 2.17 is 2.16
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Triangles and angles<br> Pls help
AleksandrR [38]

Answer:

The two bottom angles of this triangle are both represented by the same variable, in this case, an x. This means, that both of the bottom angles are equal to each other.

So we have:

32 + x + x = 180 (since the TOTAL angles of a triangle adds up to 180)

32 + 2x = 180 (combine the 2 x's together)

2x = 148 (subtract a 32 from both sides)

x = 74

Let me know if this helps!

4 0
3 years ago
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Sav [38]
The answer is D

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8 0
3 years ago
What is the sum of the measures of the measures of the exterior angles of this triangle?
mylen [45]

Answer:

The sum of the exterior angles is 360 degrees.

7 0
3 years ago
Use the distributive property to fill in the blank below.<br><br> 8 x (3+5)=( _ x 3 ) + ( _ x 5 )
olganol [36]
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5 0
3 years ago
The proportion of U.S. births that result in a birth defect is approximately 1/33 according to the Centers for Disease Control a
Dafna11 [192]

Answer:

Probability that at least 490 do not result in birth defects = 0.1076

Step-by-step explanation:

Given - The proportion of U.S. births that result in a birth defect is approximately 1/33 according to the Centers for Disease Control and Prevention (CDC). A local hospital randomly selects five births and lets the random variable X count the number not resulting in a defect. Assume the births are independent.

To find - If 500 births were observed rather than only 5, what is the approximate probability that at least 490 do not result in birth defects

Proof -

Given that,

P(birth that result in a birth defect) = 1/33

P(birth that not result in a birth defect) = 1 - 1/33 = 32/33

Now,

Given that, n = 500

X = Number of birth that does not result in birth defects

Now,

P(X ≥ 490) = \sum\limits^{500}_{x=490} {^{500} C_{x} } (\frac{32}{33} )^{x} (\frac{1}{33} )^{500-x}

                 = {^{500} C_{490} } (\frac{32}{33} )^{490} (\frac{1}{33} )^{500-490}  + .......+ {^{500} C_{500} } (\frac{32}{33} )^{500} (\frac{1}{33} )^{500-500}

                = 0.04541 + ......+0.0000002079

                = 0.1076

⇒Probability that at least 490 do not result in birth defects = 0.1076

4 0
3 years ago
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