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3 years ago

What is 48/8+5*(7-2) please help asap

Mathematics

2 answers:

maria [59]3 years ago

7 0

The answer is 31 for the problem.

Alexeev081 [22]3 years ago

4 0

Answer:

Step-by-step explanation:

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Use a table of values to estimate the value of the limit. Confirm your result graphically by graphing the function with a graphi

il63 [147K]

Hello,

$\lim_{x \to 0}\dfrac{ \sqrt{x+25}-5}{x}\\\\$

$=\lim_{x \to 0}\dfrac{(\sqrt{x+25}-5)*(\sqrt{x+25}+5)}{x*(\sqrt{x+25}+5)}\\\\$

$=\lim_{x \to 0}\dfrac{x+25-25)}{x*(\sqrt{x+25}+5)}\\\\$

$=\dfrac{1}{10}$

4 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

2 years ago

richard spends 30% of his monthly budget on rent. His total monthly is 3,000. How much does Ricard spend on rent

Alexeev081 [22]

He spends 900 on rent

6 0

3 years ago

Read 2 more answers

Find the value of x.

allsm [11]

Answer:

$3 \sqrt{5}$

Step-by-step explanation:

x is the center of the circle.

the chord of length 12 can be splitted into two parts of 6 length each.

since the radius does not change when it touches another end of the circle, we can safely calculate our radius to be the hypotenuse of the triangle formed by the length of the chord and 3.

The explanation is given in the picture

5 0

2 years ago

The diameter of Jupiter is 88,000 miles This is about 11.1 times the diameter of Earth What is the diameter of the earth

Vadim26 [7]

Assuming that Jupiter and Earth are similar
linear sale factor=(diameter of Jupiter) /(diameter of earth)=11
thus:
diameter of earth=(diameter of Jupiter)/(linear scale factor)
=88000/11
=8000 miles
The diameter of earth is 8000 miles

5 0

3 years ago