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ValentinkaMS [17]

3 years ago

11

Mrs. Gomes found that 40% of students at her high school take chemistry. She randomly surveys 12 students. What is the probabili

ty that at most 4 students have taken chemistry? Round the answer to the nearest thousandth.
mc026-1.jpg
0.008
0.438
0.562
0.665

2 answers:

sweet-ann [11.9K]3 years ago

5 0

Answer:

B

Step-by-step explanation:

JUST TOOK THE UNIT TEST REVIEW

Virty [35]3 years ago

4 0

Answer:

C

Step-by-step explanation:

C

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4 2/3 divided by 1/2

dimulka [17.4K]

Answer:

9.33333333333

Step-by-step explanation:

5 0

2 years ago

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A survey of 2541 American households discovered<br> that 64% of the households own one car.

Ipatiy [6.2K]

Answer:

1626.24

Step-by-step explanation:

You need to take the sample(2541) and divide it by the percentage so you would do 2541 x .64 =1626.24

6 0

3 years ago

The region bounded by y=(3x)^(1/2), y=3x-6, y=0

Ganezh [65]

Answer:

4.5 sq. units.

Step-by-step explanation:

The given curve is $y = (3x)^{\frac{1}{2} }$

⇒ $y^{2} = 3x$ ...... (1)

This curve passes through (0,0) point.

Now, the straight line is y = 3x - 6 ....... (2)

Now, solving (1) and (2) we get,

$y^{2} - y - 6 = 0$

⇒ (y - 3)(y + 2) = 0

⇒ y = 3 or y = -2

We will consider y = 3.

Now, y = 3x - 6 has zero at x = 2.

Therefor, the required are = $\int\limits^3_0 {(3x)^{\frac{1}{2} } } \, dx - \int\limits^3_2 {(3x - 6)} \, dx$

= $\sqrt{3} [{\frac{x^{\frac{3}{2} } }{\frac{3}{2} } }]^{3} _{0} - [\frac{3x^{2} }{2} - 6x ]^{3} _{2}$

= $[\frac{\sqrt{3}\times 2 \times 3^{\frac{3}{2} } }{3}] - [13.5 - 18 - 6 + 12]$

= 6 - 1.5

= 4.5 sq. units. (Answer)

7 0

3 years ago

Which of the following best describes the equation below?<br><br> y = |x| + 5

sleet_krkn [62]

Answer:

On a graph, you go up 5.

4 0

3 years ago

Two problems here I need solved! I need every step, so please have that with your answers!!

Free_Kalibri [48]

QUESTION 1

We want to solve,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-6x+8}$

We factor the denominator of the fraction on the right hand side to get,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-4x - 2x+8}.$

This implies
$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x(x-4) - 2(x - 4)}.$

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{(x-4)(x - 2)}$

We multiply through by LCM of
$(x-4)(x - 2)$

$(x - 2) + x(x-4) = 2$

We expand to get,

$x - 2 + {x}^{2} - 4x= 2$

We group like terms and equate everything to zero,

${x}^{2} + x - 4x - 2 - 2 = 0$

We split the middle term,

${x}^{2} + - 3x - 4 = 0$

We factor to get,

${x}^{2} + x - 4x- 4 = 0$

$x(x + 1) - 4(x + 1) = 0$

$(x + 1)(x - 4) = 0$

$x + 1 = 0 \: or \: x - 4 = 0$

$x = - 1 \: or \: x = 4$

But
$x = 4$
is not in the domain of the given equation.

It is an extraneous solution.

$\therefore \: x = - 1$
is the only solution.

QUESTION 2

$\sqrt{x+11} -x=-1$

We add x to both sides,

$\sqrt{x+11} =x-1$

We square both sides,

$x + 11 = (x - 1)^{2}$

We expand to get,

$x + 11 = {x}^{2} - 2x + 1$

This implies,

${x}^{2} - 3x - 10 = 0$

We solve this quadratic equation by factorization,

${x}^{2} - 5x + 2x - 10 = 0$

$x(x - 5) + 2(x - 5) = 0$

$(x + 2)(x - 5) = 0$

$x + 2 = 0 \: or \: x - 5 = 0$

$x = - 2 \: or \: x = 5$

But
$x = - 2$
is an extraneous solution

$\therefore \: x = 5$

7 0

3 years ago