I NEED HELP PLEASE ...ASAP!!!!!

In a clothing store there are 12 red dresses, 5 blue shirts, and 8 black hats. Write the ratios:

lisabon 2012 [21]

Answer:

B?

Step-by-step explanation:

Just guessed.

6 0

1 year ago

Data from the article "The Osteological Paradox: Problems inferring Prehistoric Health from Skeletal Samples" (Current Anthropol

iris [78.8K]

Answer:

(a) P (X < 109.78) = 0.9484.

(b) P (X < 109.78) = 0.9484.

(c) P (97 < X < 106) = 0.5328.

(d) P (X < 85.6 or X > 111.4) = 0.0369.

(e) P (X > 103) = 0.3085.

(f) P (X < 98.2) = 0.3821.

(g) P (100 < X < 124) = 0.5000.

(h) The middle 80% of all heights of 5 year old children fall between 92.31 and 107.70.

Step-by-step explanation:

It is provided that X follows a Normal distribution with mean, μ = 100 and standard deviation, σ = 6.

(a)

Compute the value of P (X > 89.2) as follows:

$P (X>89.2)=P(\frac{X-\mu}{\sigma}>\frac{89.2-100}{6})\\=P(Z>-1.80)\\=P(Z$

Thus, the value of P (X > 89.2) is 0.9641.

(b)

Compute the value of P (X < 109.78) as follows:

$P (X$

Thus, the value of P (X < 109.78) is 0.9484.

(c)

Compute the value of P (97 < X < 106) as follows;

P (97 < X < 106) = P (X < 106) - P (X < 97)

$=P(\frac{X-\mu}{\sigma}$

Thus, the value of P (97 < X < 106) is 0.5328.

(d)

Compute the value of P (X < 85.6 or X > 111.4) as follows;

P (X < 85.6 or X > 111.4) = P (X < 85.6) + P (X > 111.4)

$=P(\frac{X-\mu}{\sigma}\frac{111.4-100}{6})\\=P(Z1.9)\\=0.0082+0.0287\\=0.0369$

Thus, the value of P (X < 85.6 or X > 111.4) is 0.0369.

(e)

Compute the value of P (X > 103) as follows:

$P (X>103)=P(\frac{X-\mu}{\sigma}>\frac{103-100}{6})\\=P(Z>0.50)\\=14-P(Z$

Thus, the value of P (X > 103) is 0.3085.

(f)

Compute the value of P (X < 98.2) as follows:

$P (X$

Thus, the value of P (X < 98.2) is 0.3821.

(g)

Compute the value of P (100 < X < 124) as follows;

P (100< X < 124) = P (X < 124) - P (X < 100)

$=P(\frac{X-\mu}{\sigma}$

Thus, the value of P (100 < X < 124) is 0.5000.

(h)

Compute the value of x₁ and x₂ as follows if P (x₁ < X < x₂) = 0.80 as follows:

$P(X_{1}$

The value of z is ± 1.282.

The value of x₁ and x₂ are:

$-z=\frac{x_{1}-\mu}{\sigma} \\-1.282=\frac{x_{1}-100}{6}\\x_{1}=100-(6\times1.282)\\=92.308\\\approx92.31$ $z=\frac{x_{2}-\mu}{\sigma} \\1.282=\frac{x_{2}-100}{6}\\x_{2}=100+(6\times1.282)\\=107.692\\\approx107.70$

Thus, the middle 80% of all heights of 5 year old children fall between 92.31 and 107.70.

5 0

2 years ago

The number of cubic feet of water in a curved container can be approximated by v = 0.95h2.9. use a calculator to find the amount

Sati [7]

The first thing we must do for this case is to rewrite the expression of the volume correctly.
We have then:
v = 0.95 * h ^ 2.9
Substituting the value of h we have:
v = 0.95 * (7.5) ^ 2.9
v = 327.6432831
round to the nearest tenth:
v = 327.6 feet ^ 3
Answer:
the amount of water in the container when h = 7.5 feet is:
v = 327.6 feet ^ 3

3 0

2 years ago

Which subsets does 65/13 belong? Move the appropriate subsets into this box.

zhannawk [14.2K]

Answer:

roam

Step-by-step explanation:

free points

3 0

2 years ago

Two ships leave a harbor together, traveling on courses that have an angle of 135°40' between them. If they each travel 402 mile

mario62 [17]

Answer:

Therefore they are 734.106 miles apart.

Step-by-step explanation:

Given that ,

Two ships have a harbor together. The angle between two ships is 135°40'. Each of two ships travel 402 miles.

It forms a isosceles triangle whose two sides are 402 miles and one angle is 135°40'. Since it is isosceles triangle then other two angles of the triangle is equal.

Let ∠B= 135°40', and AB = 402 miles , BC = 402 miles

Then the distance between the ships = AC

We know

The sum of all angles = 180°

⇒∠A+∠B+∠C=180°

⇒∠A+135°40'+∠C=180°

⇒2∠A= 180°- 135°40' [ since ∠A=∠C]

⇒2∠A=44°60'

⇒∠A= 22°30'

Again we know that,

$\frac{AB}{sin\angle C}=\frac{BC}{sin \angle A}=\frac{AC}{sin \angle B}$

Taking last two ratio,

$\frac{BC}{sin \angle A}=\frac{AC}{sin \angle B}$

Putting the value of BC , AC ,∠A,∠B

$\frac{402}{sin 22^\circ30'}=\frac{AC}{sin 135^\circ40'}$

$\Rightarrow AC=\frac{402 \times sin135^\circ40'}{sin 22^\circ30'}$

≈734.106 miles

Therefore they are 734.106 miles apart.

3 0

2 years ago