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3 years ago

Find the missing value (x)

Mathematics

1 answer:

Semenov [28]3 years ago

3 0

Answer:

y=14; x=72

Step-by-step explanation:

y=(21×2)/3; x=(48×21)/14

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What is the domain and range for this graph?

Kay [80]

Domain: -∞, ∞
range: -5,∞
(the commas mean through; ex.; -5,∞..... that means negative 5 thru infinity)

8 0

4 years ago

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The weekly earnings of students in one age group are normally distributed with a standard deviation of 47 dollars. A researcher

ELEN [110]

Answer:

Option D) 340

Step-by-step explanation:

We are given the following in the question:

Alpha, α = 0.05

Population standard deviation, σ = $47

Margin of error = 5

95% Confidence Interval:

$\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}$

$\text{Margin of error} = z_{critical}\frac{\sigma}{\sqrt{n}}$

Putting the values, we get,

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

$5 = 1.96(\dfrac{47}{\sqrt{n}} )\\\\\sqrt{n} = \dfrac{47\times 1.96}{5}\\\\n = 339.443776\\\Rightarrow n \approx 340$

Thus, the correct answer is

Option D) 340

7 0

3 years ago

Help!

inessss [21]

Answer:

2 inches is the height of the box.

Step-by-step explanation:

4 0

3 years ago

Find the least common multiple (LCM) of 8y^6+ 144y^5+ 640y^4 and 2y^4 + 40y^3 + 200y^2.

Mice21 [21]

Answer:

LCM = $8y^{4}(y+ 10)^{2}(y + 8)$

Step-by-step explanation:

Making factors of $8y^{6}+ 144y^{5}+ 640y^{4}$

Taking $8y^{4}$ common:

$\Rightarrow 8y^{4} (y^{2}+ 18y+ 80)$

Using factorization method:

$\Rightarrow 8y^{4} (y^{2}+ 10y + 8y + 80)\\\Rightarrow 8y^{4} (y (y+ 10) + 8(y + 10))\\\Rightarrow 8y^{4} (y+ 10)(y + 8))\\\Rightarrow \underline{2y^{2}} \times 4y^{2} \underline{(y+ 10)}(y + 8)) ..... (1)$

Now, Making factors of $2y^{4} + 40y^{3} + 200y^{2}$

Taking $2y^{2}$ common:

$\Rightarrow 2y^{2} (y^{2}+ 20y+ 100)$

Using factorization method:

$\Rightarrow 2y^{2} (y^{2}+ 10y+ 10y+ 100)\\\Rightarrow 2y^{2} (y (y+ 10) + 10(y + 10))\\\Rightarrow \underline {2y^{2} (y+ 10)}(y + 10) ............ (2)$

The underlined parts show the Highest Common Factor(HCF).

i.e. HCF is $2y^{2} (y+ 10)$ .

We know the relation between LCM, HCF of the two numbers 'p' , 'q' and the numbers themselves as:

$HCF \times LCM = p \times q$

Using equations (1) and (2): $\Rightarrow 2y^{2} (y+ 10) \times LCM = 2y^{2} \times 4y^{2}(y+ 10)(y + 8) \times 2y^{2} (y+ 10)(y + 10)\\\Rightarrow LCM = 2y^{2} \times 4y^{2}(y+ 10)(y + 8) \times (y + 10)\\\Rightarrow LCM = 8y^{4}(y+ 10)^{2}(y + 8)$

Hence, LCM = $8y^{4}(y+ 10)^{2}(y + 8)$

5 0

3 years ago

Using the numbers -4,10, 8, 2, -3, -5 , create two expreasioms that equal 6

timama [110]

I would say take 2 from eight which equals 6

5 0

3 years ago

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