What does 1 + 9x equal?

Mathematics

1 answer:

iragen [17]3 years ago

7 0

Answer:

9x+1

Step-by-step explanation:

Let's simplify step-by-step.

1+9x

Answer:

=9x+1

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Please answer right away

Andreas93 [3]

Answer:

20 miles with an error margin of ± 8 miles

Step-by-step explanation:

The margin of error of a result is the range in which an error can vary. To find the margin of error between both distances we have to

28-12 = 16, that is, the variation of the result has a range of 16 miles. So we will look for the midpoint of both distances

(X2-X1)/2+X1=(28-12)/2+12=16/2+12=8+12=20

So from this midpoint the value can vary between 8 points below and 8 points above that would cover the difference of 16 miles that we observed at the beginning

In this way, the correct answer is 20 miles with an error margin of ± 8 miles

Done

8 0

3 years ago

Clue 1

nikklg [1K]

The number is 112
1. Three digits
2. Less than 140
3. 7 is a factor
4. Even
5. 1+1+2 = 4

7 0

2 years ago

assume that when adults with smartphones are randomly selected 15 use them in meetings or classes if 15 adult smartphones are ra

Tanzania [10]

Answer:

The probability that at least 4 of them use their smartphones is 0.1773.

Step-by-step explanation:

We are given that when adults with smartphones are randomly selected 15% use them in meetings or classes.

Also, 15 adult smartphones are randomly selected.

Let X = <em>Number of adults who use their smartphones</em>

The above situation can be represented through the binomial distribution;

$P(X = r) = \binom{n}{r}\times p^{r} \times (1-p)^{n-r} ; n = 0,1,2,3,.......$

where, n = number of trials (samples) taken = 15 adult smartphones

r = number of success = at least 4

p = probability of success which in our question is the % of adults

who use them in meetings or classes, i.e. 15%.

So, X ~ Binom(n = 15, p = 0.15)

Now, the probability that at least 4 of them use their smartphones is given by = P(X $\geq$ 4)

P(X $\geq$ 4) = 1 - P(X = 0) - P(X = 1) - P(X = 2) - P(X = 3)

= $1- \binom{15}{0}\times 0.15^{0} \times (1-0.15)^{15-0}-\binom{15}{1}\times 0.15^{1} \times (1-0.15)^{15-1}-\binom{15}{2}\times 0.15^{2} \times (1-0.15)^{15-2}-\binom{15}{3}\times 0.15^{3} \times (1-0.15)^{15-3}$