Are these answers correct

Given that 196k is a perfect cube , find the smallest possible value of k

kaheart [24]

Factor 196k
196k=2*2*7*7*k
we need to get something like x^3
hmm
we right now have
(2*7)^2
if we multiply that by (2*7)^1, we can beg (2*7)^3,
therefor k=2*7=14

tha tnumber is 14

or

if k can be any number
k can be 1/196, to make it 1, and 1 is a perfect cube

or

my bro had an epic idea: what if k is negative
the cube root of a negative number is negative
so the smallest one would be negative infinity or something

I think the answer you're looking for is k=14

3 0

4 years ago

Please help me . factor. X^2+22x+40

Elanso [62]

(x + 2) (x + 20) is the answer.

7 0

4 years ago

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Casey has a loan for $400 at a rate of 13% annually. If the interest is not compounded, how much interest will he pay in 3 years

Drupady [299]

$400 * 0.13 * 3 = $156

answer is <span>C) $156.00 </span>

4 0

4 years ago

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Question Help Suppose that the lifetimes of light bulbs are approximately normally distributed, with a mean of 5656 hours and a

koban [17]

Answer:

a)3.438% of the light bulbs will last more than 6262 hours.

b)11.31% of the light bulbs will last 5252 hours or less.

c) 23.655% of the light bulbs are going to last between 5858 and 6262 hours.

d) 0.12% of the light bulbs will last 4646 hours or less.

Step-by-step explanation:

Normally distributed problems can be solved by the z-score formula:

On a normaly distributed set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a value X is given by:

$Z = \frac{X - \mu}{\sigma}$

After we find the value of Z, we look into the z-score table and find the equivalent p-value of this score. This is the probability that a score will be LOWER than the value of X.

In this problem, we have that:

The lifetimes of light bulbs are approximately normally distributed, with a mean of 5656 hours and a standard deviation of 333.3 hours.

So $\mu = 5656, \sigma = 333.3$

(a) What proportion of light bulbs will last more than 6262 hours?

The pvalue of the z-score of $X = 6262$ is the proportion of light bulbs that will last less than 6262. Subtracting 100% by this value, we find the proportion of light bulbs that will last more than 6262 hours.