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Oksi-84 [34.3K]

2 years ago

7

ℎ ℎ 22 2 22 2− 3 +10

2 answers:

Alecsey [184]2 years ago

3 0

Answer:

wouldn't the answer be 9

Step-by-step explanation:

murzikaleks [220]2 years ago

3 0

Answer:

22h^2 + 33 (added) . h⋅h⋅22,2,22,2−3+10 (subtracted)

Step-by-step explanation:

I don't understand what you want so I will list lots of answers hopefully it helps. Please message under my answer for more solutions so I can help you but I hope this started to help you.

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A set of average city temperatures in July are normally distributed

Archy [21]

Answer:

a) 59.87%

Step-by-step explanation:

Given that the mean temperature (μ) = 23.5°C and the standard deviation (σ) is 2 °C.

The z score is used in statistics to measure number of standard deviation by which the raw score is above or below the mean value, It is calculated using the equation:

$z=\frac{x-\mu}{\sigma}$

The average temperature of the City of Van Nuys is 23 degrees. The z score is

$z=\frac{x-\mu}{\sigma}=\frac{23-23,5}{2} =-0.25$

P(x > 23) = P(z > -0.25) = 1 - P(z < -0.25) = 1 - 0.4013 = 0.5987 = 59.87%

3 0

2 years ago

X+y+(y-2)=60,1/2(x)(y-2)=120 what is x and y

Rudiy27

I think is this hopefully this helps

5 0

2 years ago

Read 2 more answers

Pls answer my question it is urgent

yaroslaw [1]

Answer: rs 400 and 500

Step-by-step explanation:

3 0

2 years ago

8. A farm packs 400 kg of strawberries so

Simora [160]

Answer:

I think the answer is 10 full 6Kg boxes

my first answer if you saw it was miscalculated, sorry.

3 0

3 years ago

Read 2 more answers

1011+111 in binary form

Ann [662]

Answer:

10010

Step-by-step explanation:

$1011=1(2)^3+0(2)^2+1(2)^1+1(2)^0$

$111=1(2)^2+1(2)^1+1(2)^0$

So $1011+111$ gives us:

$1(2)^3+0(2)^2+1(2)^1+1(2)^0$

$+$

$1(2)^2+1(2)^1+1(2)^0$

-----------------------------------------------------

Combine like terms:

$1(2)^3+(0+1)(2)^2+(1+1)(2)^1+(1+1)(2)^0$

$1(2)^3+1(2)^2+(2)(2)^1+(2)(2)^0$

We aren't allowed to have a coefficient bigger than 1.

I'm going to replace $2^0$ with 1 and $2$ with $(2)^1$ :

$1(2)^3+1(2)^2+(2)^2+(2)^1(1)$

I want a $2^0$ number:

$1(2)^3+1(2)^2+1(2)^2+1(2)^1+0(2)^0$

Combine like terms:

$1(2)^3+2(2)^2+1(2)^1+0(2)^0$

$2(2)^2=2^3$ :

$1(2)^3+2^3+1(2)^1+0(2)^0$

Combine like terms:

$2(2)^3+1(2)^1+0(2)^0$

We can rewrite the first term by law of exponents:

$2^4+1(2)^1+0(2)^0$

$1(2)^4+1(2)^1+0(2)^0$

So the binary form is:

$10010$

Maybe you like this way more:

Keep in mind 1+1=10 and that 1+1+1=11:

Setup:

1 0 1 1

+ 1 1 1

------------------------------

(1) (1) (1)

1 0 1 1

+ 1 1 1

------------------------------

1 0 0 1 0

I had to do some carry over with my 1+1=10 and 1+1+1=11.

8 0

3 years ago