A 20-ounce bottle of lotion costs $3.95. How much does each ounce cost?\

Answer:

90

Step-by-step explanation:

I'm not sure about the answer

since we know the diameter is 50, then its radius is half that or 25.

[tex]\textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=25 \end{cases}\implies \begin{array}{llll} V=\cfrac{4\pi (25)^3}{3}\implies V=\cfrac{62500\pi }{3} \\\\\\ V\approx 65449.85~cm^3 \end{array}[/tex]

Answer:

800ml

Step-by-step explanation:

solution A= 2% alcohol

solution B=7% alcohol

1200ml=1.2L

1200ml=1200cm^3

X= solution B

solution A + solution B=(1200+x)cm^3

% alcohol= amount of alcohol/total solution(AorB)×100

☆in 1200ml of solution A there's

0.02x1200

2x12

=24ml of alcohol (in 1200ml of solution A)

0.07x X

=0.07Xml of alcohol (in some ml of solution B)

(24+0.07X)ml of alcohol in = solution A and solution B

solution A + solution B= 1200+X

0.04 (1200+X)=24+0.07X

48+0.04X=24+0.07X

48-24=0.07X-0.04X

24=0.03X

X=800ml

[tex]\text{L.H.S}\\\\=4\sin(-x) \cos(-x)\\\\=-4\sin x \cos x\\\\=-2 \cdot 2 \sin x \cos x\\\\=-2 \sin 2x\\\\=\text{R.H.S}\\\\\text{The identity is true.}[/tex]

Answer:

x=2

y=2

Step-by-step explanation:

x+3y=8

- x-4y=-6

7y=14

y=14/7

y=2

x+3(2)=8

x+6=8

x=8-6

x=2

y=2

x=2

Answer:

[tex]1,\!848[/tex].

Step-by-step explanation:

There are two disjoint sets of ways to choose a lineup as required:

Assume that none of Alicia, Amanda, or Anna is to be selected. This lineup of [tex]6[/tex] would then need to be selected from a set of [tex]14 - 3 = 11[/tex] players (which excludes Alicia, Amanda, and Anna.)

The number of ways of selecting (without order) [tex]6[/tex] items out of a set of [tex]11[/tex] (distinct) items is equal to the combination:

[tex]\begin{aligned}\begin{pmatrix}11 \\ 6\end{pmatrix} &= \frac{11!}{(6!)\, (11 - 6)!} \\ &= \frac{11!}{6! \times 5!}\end{aligned}[/tex].

Assume that Alicia is selected, but neither Amanda nor Anna is selected. The other [tex]6 - 1 = 5[/tex] players in this lineup would then need to be selected from a set of [tex]14 - 1 - 2 = 11[/tex] players. (This set of [tex]11[/tex] excludes Alicia, Amanda, and Anna.)

The number of ways to select [tex]5[/tex] items from a set of [tex]11[/tex] items is:

[tex]\begin{aligned}\begin{pmatrix}11 \\ 5\end{pmatrix} &= \frac{11!}{(5!)\, (11 - 5)!} \\ &= \frac{11!}{5! \times 6!} \\ &= \frac{11!}{6! \times 5!}\end{aligned}[/tex].

Similarly, there would be another set of [tex](11!) / (6! \times 5!)[/tex] distinct ways to select the lineup if Amanda is selected, but neither Alicia nor Anna is.

Likewise, the number of ways to select the lineup with Anna but neither Amanda nor Alicia would also be [tex](11!) / (6! \times 5!)[/tex].

These sets of configurations for the lineup are pairwise disjoint from one another. Thus, the total number of ways to select this lineup would be:

[tex]\begin{aligned}& \begin{pmatrix}11 \\ 6 \end{pmatrix} + 3 \times \begin{pmatrix}11 \\ 5 \end{pmatrix} \\ =\; & \frac{11!}{6! \times 5!} + 3 \times \frac{11!}{6! \times 5!} \\ =\; & \frac{4 \times 11!}{6! \times 5!} \\ =\; & \frac{4 \times 11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2} \\ =\; & \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 3 \times 2} \\ =\; & 1,\!848\end{aligned}[/tex].