Find the IQR of the data in the box plot below

ddd [48]

Answer:

$\frac{x - 10}{8}$

Step-by-step explanation:

$\frac{3x}{8} - \frac{x + 5}{4}$

The common denominator of both fractions is 8. Therefore,, divide the denominator of each fraction, then multiply what you get by the numerator of each fraction.

Thus:

$\frac{1(3x) - 2(x + 5)}{8}$

$\frac{3x - 2x - 10}{8}$

$\frac{x - 10}{8}$

5 0

3 years ago

Suppose there are 4 defective batteries in a drawer with 10 batteries in it. A sample of 3 is taken at random without replacemen

SSSSS [86.1K]

Answer:

a.) 0.5

b.) 0.66

c.) 0.83

Step-by-step explanation:

As given,

Total Number of Batteries in the drawer = 10

Total Number of defective Batteries in the drawer = 4

⇒Total Number of non - defective Batteries in the drawer = 10 - 4 = 6

Now,

As, a sample of 3 is taken at random without replacement.

a.)

Getting exactly one defective battery means -

1 - from defective battery

2 - from non-defective battery

So,

Getting exactly 1 defective battery = ⁴C₁ × ⁶C₂ = $\frac{4!}{1! (4 - 1 )!}$ × $\frac{6!}{2! (6 - 2 )!}$

= $\frac{4!}{(3)!}$ × $\frac{6!}{2! (4)!}$

= $\frac{4.3!}{(3)!}$ × $\frac{6.5.4!}{2! (4)!}$

= $4$ × $\frac{6.5}{2.1! }$

= $4$ × $15$ = 60

Total Number of possibility = ¹⁰C₃ = $\frac{10!}{3! (10-3)!}$

= $\frac{10!}{3! (7)!}$

= $\frac{10.9.8.7!}{3! (7)!}$

= $\frac{10.9.8}{3.2.1!}$

= 120

So, probability = $\frac{60}{120} = \frac{1}{2} = 0.5$

b.)

at most one defective battery :

⇒either the defective battery is 1 or 0

If the defective battery is 1 , then 2 non defective

Possibility = ⁴C₁ × ⁶C₂ = 60

If the defective battery is 0 , then 3 non defective

Possibility = ⁴C₀ × ⁶C₃

= $\frac{4!}{0! (4 - 0)!}$ × $\frac{6!}{3! (6 - 3)!}$

= $\frac{4!}{(4)!}$ × $\frac{6!}{3! (3)!}$

= 1 × $\frac{6.5.4.3!}{3.2.1! (3)!}$

= 1× $\frac{6.5.4}{3.2.1! }$

= 1 × 20 = 20

getting at most 1 defective battery = 60 + 20 = 80

Probability = $\frac{80}{120} = \frac{8}{12} = 0.66$

c.)

at least one defective battery :

⇒either the defective battery is 1 or 2 or 3

If the defective battery is 1 , then 2 non defective

Possibility = ⁴C₁ × ⁶C₂ = 60

If the defective battery is 2 , then 1 non defective

Possibility = ⁴C₂ × ⁶C₁

= $\frac{4!}{2! (4 - 2)!}$ × $\frac{6!}{1! (6 - 1)!}$

= $\frac{4!}{2! (2)!}$ × $\frac{6!}{1! (5)!}$

= $\frac{4.3.2!}{2! (2)!}$ × $\frac{6.5!}{1! (5)!}$

= $\frac{4.3}{2.1!}$ × $\frac{6}{1}$

= 6 × 6 = 36

If the defective battery is 3 , then 0 non defective

Possibility = ⁴C₃ × ⁶C₀

= $\frac{4!}{3! (4 - 3)!}$ × $\frac{6!}{0! (6 - 0)!}$

= $\frac{4!}{3! (1)!}$ × $\frac{6!}{(6)!}$

= $\frac{4.3!}{3!}$ × 1

= 4×1 = 4

getting at most 1 defective battery = 60 + 36 + 4 = 100

Probability = $\frac{100}{120} = \frac{10}{12} = 0.83$

3 0

3 years ago

Adina has a rectangular garden that measures 9 m wide by 13 m long. She wants to

MrRissso [65]

9*13=117
12*16=192

We increased three meters to both the length and width of the garden.

6 0

4 years ago

Read 2 more answers

Rosita earned $325 for working 25 hours last week. What is her hourly rate?

miskamm [114]

Answer:

13

Step-by-step explanation: 325/25=13

8 0

3 years ago

Read 2 more answers

A triangular pyramid is cut by a plane. In which case is the cross section a quadrilateral?

Pie

Answer:

(D) is the right answer

Step-by-step explanation:

The cross section will be a quadrilateral when the intersecting plane forms a closed four sided polygon along with the pyramid.

A triangular pyramid has four faces. So, to form four straight lines and therefore a quadrilateral, the intersecting plane must intersect all four faces.

A. In this case the plane will only intersect one side of the pyramid so four sided polygon can't be formed.

B. It will only intersect the two sides of the pyramid but the cross section will not be a quadrilateral.

C. In this case the plane will intersect three sides of the pyramid so the cross section will be a triangle not a quadrilateral.

D. In this case the plane intersects all the four faces forming a closed polygon of 4 straight lines so the cross section will be a quadrilateral.

So, the Correct Option is (D) : When the plane intersects the base and all three lateral faces of the pyramid, the cross section of the triangular pyramid will be a quadrilateral.

8 0

3 years ago