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3 years ago

Help me get this rightttt guys

Mathematics

1 answer:

kirill [66]3 years ago

5 0

Answer:

3000=3×10^3

300=3×10^2

30=3×10^1

3=3×10^0

0.3=3×10^-1

0.03=3×10^-2

0.003=3×10^-3

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The ages of three siblings are consecutive integers. The sum of their ages is 42. Calculate their ages. *

stealth61 [152]

Answer:

13, 14, 15

Step-by-step explanation:

Let x = age of one of the siblings

Since the age of the siblings are consecutive integers, the age of the next sibling would be x + 1 and the age of the third sibling would be x + 2

x +( x +1) + (x +2) = 42

3x + 3 = 42

collect like terms and solve for x

x = 13

The age of the first sibling is 13

The age of the second sibling is x + 1 = 13 + 1 = 14

The age of the third sibling is x + 2 = 13 + 2 = 15

8 0

3 years ago

4) Which expression is equivalent to 9(9m + 3t) ? 6.EE.A.3"

horrorfan [7]

The answer is D; could you make me the brainiest? Because the answer is 81m+27t

4 0

3 years ago

Determine length of RT

dalvyx [7]

Answer:

Step-by-step explanation:

6 0

3 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)

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

= $1- (1\times 1\times 0.85^{15})-(15\times 0.15^{1} \times 0.85^{14})-(105 \times 0.15^{2} \times 0.85^{13})-(455 \times 0.15^{3} \times 0.85^{12})$

= <u>0.1773</u>

3 0

4 years ago

Find f (x )if f apostrophe apostrophe (x )equals x to the power of negative 2 end exponent comma space x greater than 0 comma sp

viva [34]

Step-by-step explanation:

Given $g'(x) = g^{-2}, x> 0\ g(1) = 0 \ and \ g(2) = 0$

To get g(x), we will have to integrate g'(x)

$g(x) = \int\limits {x^{-2}} \, dx\\ g(x) = \frac{x^{-2+1}}{-2+1} + C\\ g(x) = -x^{-1} + C\\g(x) = -x^{-1} + C\\$

If g(1) = 0, this means at x = 1, g(x) = 0

0 = -1⁻¹ + C

C= 1

Substitute C = 1 into the function

g(x) = -x⁻¹ + 1

If g(2) = 0, this means at x = 2, g(x) = 0

0 = -2⁻¹ + C

C= 2⁻¹

C = 1/2

Substitute C = 2 into the function

g(x) = -x⁻¹ + 1/2

3 0

3 years ago