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3 years ago

What is (3/4)³ in fraction form

Mathematics

1 answer:

suter [353]3 years ago

6 0

27/64 . because you do 3/4 to the third power (3x) which gives you that

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4 7/12 + 2 1/12 what is the estimate answer

Pani-rosa [81]

$4 \frac{7}{12} = \frac{12*4+7}{12} = \frac{55}{12} \\ \\ 2 \frac{1}{12} = \frac{2*12+1}{12} = \frac{25}{12} \\ \\ \frac{55}{12} + \frac{25}{12} = \frac{80}{12} \\ \\ \frac{80 :4}{4:4} = \frac{20}{3}$

3 0

4 years ago

Read 2 more answers

The region bounded by the given curve is rotated about the specified axis. find the volume v of the resulting solid by any metho

Blababa [14]

$x^2+(y-2)^2=4\implies y=2\pm\sqrt{4-x^2}$

describes a circle of radius 2 centered at (0,2). A revolution of this region about the y-axis would give you a sphere of radius 2, so the volume would be $\dfrac43\pi\times2^3=\dfrac{32}3\pi$ .

4 0

3 years ago

If f (x) = -9x - 9 and g (x) = Vx - 9, what is (f ° g) (10)?

Temka [501]

Answer: $(f \circ g) (10)= -18\ .$

Step-by-step explanation:

Given: $f (x) = -9x - 9$ and $g (x) = \sqrt{x - 9}$

To find : (f o g) (10)

For this we first find (f o g) (x)= $f(g(x))$

$=f(\sqrt{x-9})\\\\=-9(\sqrt{x-9})-9$

Now,

$(f \circ g) (10)=-9(\sqrt{10-9})-9\\\\=-9\sqrt{1}-9\\\\=-9-9=-18$

Hence, the value of $(f \circ g) (10)= -18\ .$

8 0

3 years ago

birth weights in norway are normally distributed with a mean of 3570 g and a standard deviation of 500g. if the hospital officia

Svetradugi [14.3K]

Answer:

2630 g

Step-by-step explanation:

From the given information:

Given that:

mean (μ) = 3750 g

Standard deviation (σ) = 500

Suppose the hospital officials demand special treatment with a percentage of lightest 3% (0.03) for newborn babies;

Then, the weight of birth that differentiates the babies that needed special treatment from those that do not can be computed as follows;

P(Z < z₁) = 0.03

Using the Excel Formula; =NORMSINV(0.03) = -1.88

z₁ = - 1.88

Using the test statistics z₁ formula:

$z_1 = \dfrac{X-\mu}{\sigma}$

$-1.88 = \dfrac{X-3570}{500}$

By cross multiply, we have:

-1.88 × 500 = X - 3570

-940 = X - 3570

-X = -3570 + 940

-X = -2630

X = 2630 g

Hence, 2630 g is the required weight of birth that differentiates the babies that needed special treatment from those that do not

6 0

3 years ago

(Hypothetical.) Suppose a certain person's reaction time, in seconds, for pressing a button on a visual cue has the following cu

Sonbull [250]

Answer:

the probability the person's reaction time will be between 0.9 and 1.1 seconds is 0.0378

Step-by-step explanation:

Given the data in the question;

the cumulative distribution function F(x) = $1 - \frac{1}{(x + 1)^3}$ ; $x > \theta$

probability the person's reaction time will be between 0.9 and 1.1 seconds

P( 0.9 < x < 1.1 ) = P( x ≤ 1.1 ) - P( x ≤ 0.9 )

P( 0.9 < x < 1.1 ) = F(1.1) - F(0.9)

= [ $1 - \frac{1}{(x + 1)^3}$ ] - [ $1 - \frac{1}{(x + 1)^3}$ ]

we substitute

= [ $1 - \frac{1}{(1.1 + 1)^3}$ ] - [ $1 - \frac{1}{(0.9 + 1)^3}$ ]

= [ $1 - \frac{1}{(2.1)^3}$ ] - [ $1 - \frac{1}{(1.9)^3}$ ]

= [ $1 - \frac{1}{(9.261)}$ ] - [ $1 - \frac{1}{(6.859)}$ ]

= [ 1 - 0.1079796998 ] - [ 1 - 0.1457938 ]

= 0.8920203 - 0.8542062

= 0.0378

Therefore, the probability the person's reaction time will be between 0.9 and 1.1 seconds is 0.0378

5 0

3 years ago