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3 years ago

(24 + 4) ÷ 2 = (24 ÷ 2) + ( __ ÷ 2) i dont know this

Mathematics

2 answers:

kupik [55]3 years ago

6 0

Answer:

The answer is 4.

Step-by-step explanation:

(24 + 4) ÷ 2 = (24 ÷ 2) + (x ÷2)

=28 ÷ 2 = 12 + (x÷2)

14 = 12 + (x ÷ 2)

14 - 12 = x ÷ 2

2 = x ÷ 2

2 × 2 = x

x = 4

tensa zangetsu [6.8K]3 years ago

3 0

Answer:

Step-by-step explanation:

(24+4) ÷ 2 = (24 ÷ 2) + (4 ÷ 2)

28÷2 = 14

Check:

24÷2 = 12

4÷2 = 2

12+2=14

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PLZ HELP ASAP!! ILL GIVE BRAINLEST

11111nata11111 [884]

Answer:

(-5,-4)

Step-by-step explanation:

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3 years ago

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HELPPP!!!!! Plzzzz!!

Lina20 [59]

Given:

Cards labelled 1, 3, 5, 6, 8 and 9.

A card is drawn and not replaced. Then a second card is drawn at random.

To find:

The probability of drawing 2 even numbers.

Solution:

We have,

Even number cards = 6, 8

Odd numbers cards = 1, 3, 5, 9

Total cards = 1, 3, 5, 6, 8 and 9

Number of even cards = 2

Number of total cards = 6

So, the probability of getting an even card in first draw is:

$P_1=\dfrac{\text{Number of even cards}}{\text{Number of total cards}}$

$P_1=\dfrac{2}{6}$

$P_1=\dfrac{1}{3}$

Now,

Number of remaining even cards = 1

Number of remaining cards = 5

So, the probability of getting an even card in second draw is:

$P_2=\dfrac{\text{Number of remaining even cards}}{\text{Number of remaining total cards}}$

$P_2=\dfrac{1}{5}$

The probability of drawing 2 even numbers is:

$P=P_1\times P_2$

$P=\dfrac{1}{3}\times \dfrac{1}{5}$

$P=\dfrac{1}{15}$

Therefore, the probability of drawing 2 even numbers is $\dfrac{1}{15}$ . Hence, the correct option is (b).

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3 years ago

Find the length of side x in simplest radical form with a rational denominator.<br> X

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2 years ago

Solve for x in the equation 2x^2+3x-7=x^2+5x+39

Shalnov [3]

Hey there, hope I can help!

$\mathrm{Subtract\:}x^2+5x+39\mathrm{\:from\:both\:sides}$
$2x^2+3x-7-\left(x^2+5x+39\right)=x^2+5x+39-\left(x^2+5x+39\right)$

Assuming you know how to simplify this, I will not show the steps but can add them later on upon request
$x^2-2x-46=0$

Lets use the quadratic formula now
$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$
$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:} a=1,\:b=-2,\:c=-46: x_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:1\left(-46\right)}}{2\cdot \:1}$

$\frac{-\left(-2\right)+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \ \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \frac{2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}$

Multiply the numbers 2 * 1 = 2
$\frac{2+\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}$

$2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \ \sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}$

$\mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \sqrt{\left(-2\right)^2+1\cdot \:4\cdot \:46} \ \textgreater \ \left(-2\right)^2=2^2, 2^2 = 4$

$\mathrm{Multiply\:the\:numbers:}\:4\cdot \:1\cdot \:46=184 \ \textgreater \ \sqrt{4+184} \ \textgreater \ \sqrt{188} \ \textgreater \ 2 + \sqrt{188}$
$\frac{2+\sqrt{188}}{2} \ \textgreater \ Prime\;factorize\;188 \ \textgreater \ 2^2\cdot \:47 \ \textgreater \ \sqrt{2^2\cdot \:47}$

$\mathrm{Apply\:radical\:rule}: \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b} \ \textgreater \ \sqrt{47}\sqrt{2^2}$

$\mathrm{Apply\:radical\:rule}: \sqrt[n]{a^n}=a \ \textgreater \ \sqrt{2^2}=2 \ \textgreater \ 2\sqrt{47} \ \textgreater \ \frac{2+2\sqrt{47}}{2}$

$Factor\;2+2\sqrt{47} \ \textgreater \ Rewrite\;as\;1\cdot \:2+2\sqrt{47}$
$\mathrm{Factor\:out\:common\:term\:}2 \ \textgreater \ 2\left(1+\sqrt{47}\right) \ \textgreater \ \frac{2\left(1+\sqrt{47}\right)}{2}$

$\mathrm{Divide\:the\:numbers:}\:\frac{2}{2}=1 \ \textgreater \ 1+\sqrt{47}$

Moving on, I will do the second part excluding the extra details that I had shown previously as from the first portion of the quadratic you can easily see what to do for the second part.

$\frac{-\left(-2\right)-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \ \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \frac{2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}$

$\frac{2-\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}$

$2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \ 2-\sqrt{188} \ \textgreater \ \frac{2-\sqrt{188}}{2}$

$\sqrt{188} = 2\sqrt{47} \ \textgreater \ \frac{2-2\sqrt{47}}{2}$

$2-2\sqrt{47} \ \textgreater \ 2\left(1-\sqrt{47}\right) \ \textgreater \ \frac{2\left(1-\sqrt{47}\right)}{2} \ \textgreater \ 1-\sqrt{47}$

Therefore our final solutions are
$x=1+\sqrt{47},\:x=1-\sqrt{47}$

Hope this helps!

8 0

3 years ago

Read 2 more answers

The length of human pregnancies from conception to birth varies according to an approximately normal distribution with a mean of

Varvara68 [4.7K]

Answer:

A) 68.33%

B) (234, 298)

Step-by-step explanation:

We have that the mean is 266 days (m) and the standard deviation is 16 days (sd), so we are asked:

A. P (250 x < 282)

P ((x1 - m) / sd < x < (x2 - m) / sd)

P ((250 - 266) / 16 < x < (282 - 266) / 16)

P (- 1 < z < 1)

P (z < 1) - P (-1 < z)

If we look in the normal distribution table we have to:

P (-1 < z) = 0.1587

P (z < 1) = 0.8413

replacing

0.8413 - 0.1587 = 0.6833

The percentage of pregnancies last between 250 and 282 days is 68.33%

B. We apply the experimental formula of 68-95-99.7

For middle 95% it is:

(m - 2 * sd, m + 2 * sd)

Thus,

m - 2 * sd <x <m + 2 * sd

we replace

266 - 2 * 16 <x <266 + 2 * 16

234 <x <298

That is, the interval would be (234, 298)

6 0

3 years ago