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Katen [24]
3 years ago
14

Find three consecutive integers whose sum is 21

Mathematics
2 answers:
stira [4]3 years ago
5 0

Answer:

6 + 7 + 8 = 21

Step-by-step explanation:

6, 7, and 8 are consecutive integers.

6 + 7 is 13. 13 + 8 = 21. So, 6 + 7 + 8 are consecutive integers that equal 21.

Flauer [41]3 years ago
3 0

Answer:

6+7+8= 21

Step-by-step explanation:

6,7,8 follow consecutively. and if you add them, the sum is 21

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Identify the slope and y intercept of the line having equation y=-8x-15
Harrizon [31]

Answer: the slope would be a negative slope with -15 as the y intercept.

Step-by-step explanation:

3 0
3 years ago
QUESTION: Determine if the statement is true or false? if false give an counterexample.
Jobisdone [24]

Answer:

  • False

Step-by-step explanation:

<u>Actual question is:</u>

  • logf 4(3d) + log4 1 = log4 3d

<u>Solution:</u>

  • logf 4 + logf (3d)  + 0 = log4 (3d)
  • logf 4 + logf (3d) = log4 (3d)

<u>If we assume f = 4, then</u>

  • log4 4 + log4 (3d) = log4 (3d)
  • log4 4 = 0

but log4 4 = 1

  • 1 = 0 is impossible

Therefore the statement is False

5 0
3 years ago
Write the numbers from the least to the greatest
Anvisha [2.4K]

I'm pretty sure the answer is b. because obviously 1.2 in is the least and the second least is the square root of 3.

8 0
3 years ago
What is the slope - intercept form of the linear equation 2X+3y=6?
coldgirl [10]
The answers is Y=2X-2
4 0
3 years ago
From experience, it is known that on average 10% of welds performed by a particular welder are defective. if this welder is requ
bulgar [2K]
Binomial distribution can be used because the situation satisfies all the following conditions:1. Number of trials is known and remains constant (n)2. Each trial is Bernoulli (i.e. exactly two possible outcomes) (success/failure)3. Probability is known and remains constant throughout the trials (p)4. All trials are random and independent of the othersThe number of successes, x, is then given byP(x)=C(n,x)p^x(1-p)^{n-x}whereC(n,x)=\frac{n!}{x!(n-x)!}
Here we're given
p=0.10  [ success = defective ]
n=3

(a) x=0
P(x)=C(n,x)p^x(1-p)^{n-x}
=C(3,0)0.1^0(1-0.1)^{3-0}
=1(1)(0.729)
=0.729

(b) x=2
P(x)=C(n,x)p^x(1-p)^{n-x}
=C(3,2)0.1^2(1-0.1)^{3-2}
=3(0.01)(0.9)
=0.027

(c) x &ge; 2
P(x)=\sum_{x=2}^3C(n,x)p^x(1-p)^{n-x}
=P(2)+P(3)
=C(n,2)p^2(1-p)^{n-2}+C(n,3)p^3(1-p)^{n-3}
=C(3,2)0.1^2(1-0.1)^{3-2}+C(3,3)0.1^3(1-0.1)^{3-3}
=3(0.01)(0.9)+1(0.001)1
=0.027+0.001
=0.028


8 0
3 years ago
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