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skelet666 [1.2K]
3 years ago
8

Round 23.46 to the nearest whole number.

Mathematics
2 answers:
Darina [25.2K]3 years ago
8 0
It would be rounded to 23 because if it is 4 and below it rounds down 5-9 rounds up.
Fiesta28 [93]3 years ago
4 0
It's 23 because the tenths place is 4 if it were 5 it woulb be 24
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What is 0.96 as a fraction in simplest form?
taurus [48]
I agree the answer is 24/25 :)
6 0
3 years ago
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Suppose you are going to form a committee of students and faculty. You have 10 total students and 11 total faculty to pick from.
zlopas [31]

Answer:

0.189

Step-by-step explanation:

We are using combination to solve for this.

C(n,r) = nCr = n!/r! (n - r)!

We have 10 total students and 11 total faculty = 21 people

Step 1

Selecting 3 students out of 10 total students

= 10× 9× 8×7×6×5×4×3×2×1/ 3×2×1 × 7×6×5×4×3×2×1/

= 120

Step 2

Select 6 faculty out of 11 faculty

11C6 = 11!/6!(11 - 6)!

= 11!/6! × 5!

= 11×10× 9× 8×7×6×5×4×3×2×1/6×5×4×3×2×1 × 5×4×3×2×1/

= 462

Step 3

Selecting 9 people out of 21 people to form a committee

21C9 = 21!/9!(21 - 9)!

= 21!/9! × 12!

= 293,930

Step 4

The fourth and final step of having a committee of 9 people =

(120 × 462)/293,930

= 55440/293,930

= 0.1886163372

Rounded to 3 decimal places = 0.189

Therefore, the probability that you select 3 students and 6 faculty = 0.189

3 0
3 years ago
25(y+1)-x^2(y+1) i don't know the answer
natita [175]
It seems like you've begun to group and factored out the GCF.

Because the inner binomials (the ones in parentheses) are the same, the next step is to rewrite the equation like so:

(25-x^2)(y+1)

From here, we can further factor (25-x^2) since they are both perfect squares.

(5+x)(5-x)(y+1)

No further terms can be factored so that is the final answer. Hope this helps!
7 0
3 years ago
A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 29 ft/s. Its height
Ilia_Sergeevich [38]

Answer:

\overline{v}_{@\Delta t=0.01s}=-15.22ft/s, \overline{v}_{@\Delta t=0.005s}=-15.11ft/s, \overline{v}_{@\Delta t=0.002s}=-15.044ft/s, \overline{v}_{@\Delta t=0.001s}=-15.022ft/s

Step-by-step explanation:

Now, in order to solve this problem, we need to use the average velocity formula:

\overline{v}=\frac{y_{f}-y_{0}}{t_{f}-t_{0}}

From this point on, you have two possibilities, either you find each individual y_{f}, y_{0}, t_{f}, t_{0} and input them into the formula, or you find a formula you can use to directly input the change of times. I'll take the second approach.

We know that:

t_{f}-t_{0}=\Delta t

and we also know that:

t_{f}=t_{0}+\Delta t

in order to find the final position, we can substitute this final time into the function, so we get:

y_{f}=29(t_{0}+\Delta t)-22(t_{0}+\Delta t)^{2}

so we can rewrite our formula as:

\overline{v}=\frac{29(t_{0}+\Delta t)-22(t_{0}+\Delta t)^{2}-y_{0}}{\Delta t}

y_{0} will always be the same, so we can start by calculating that, we take the provided function ans evaluate it for t=1s, so we get:

y_{0}=29t-22t^{2}

y_{0}=29(1)-22(1)^{2}

y_{0}=7ft

we can substitute it into our average velocity equation:

\overline{v}=\frac{29(t_{0}+\Delta t)-22(t_{0}+\Delta t)^{2}-7}{\Delta t}

and we also know that the initil time will always be 1, so we can substitute it as well.

\overline{v}=\frac{29(1+\Delta t)-22(1+\Delta t)^{2}-7}{\Delta t}

so we can now simplify our formula by expanding the numerator:

\overline{v}=\frac{29+29\Delta t-22(1+2\Delta t+\Delta t^{2})-7}{\Delta t}

\overline{v}=\frac{29+29\Delta t-22-44\Delta t-22\Delta t^{2}-7}{\Delta t}

we can now simplify this to:

\overline{v}=\frac{-15\Delta t-22\Delta t^{2}}{\Delta t}

Now we can factor Δt to get:

\overline{v}=\frac{\Delta t(-15-22\Delta t)}{\Delta t}

and simplify

\overline{v}=-15-22\Delta t

Which is the equation that will represent the average speed of the ball. So now we can substitute each period into our equation so we get:

\overline{v}_{@\Delta t=0.01s}=-15-22(0.01)=-15.22ft/s

\overline{v}_{@\Delta t=0.005s}=-15-22(0.005)=-15.11ft/s

\overline{v}_{@\Delta t=0.002s}=-15-22(0.002)=-15.044ft/s

\overline{v}_{@\Delta t=0.001s}=-15-22(0.001)=-15.022ft/s

5 0
3 years ago
In what quadrant is the terminal side<br> of -323°?
dexar [7]

Answer:

Quadrant 4 i believe plz tell me if im wrong

Step-by-step explanatio

6 0
3 years ago
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