Can someone help me with this problem

Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).

Free_Kalibri [48]

Answer:

4/7

Step-by-step explanation:

mark me brainliest ..............

6 0

2 years ago

First you to find the worksheet and download it<br> plase I need help

Goshia [24]

Answer:

a) The horizontal asymptote is y = 0

The y-intercept is (0, 9)

b) The horizontal asymptote is y = 0

The y-intercept is (0, 5)

c) The horizontal asymptote is y = 3

The y-intercept is (0, 4)

d) The horizontal asymptote is y = 3

The y-intercept is (0, 4)

e) The horizontal asymptote is y = -1

The y-intercept is (0, 7)

The x-intercept is (-3, 0)

f) The asymptote is y = 2

The y-intercept is (0, 6)

Step-by-step explanation:

a) f(x) = $3^{x + 2}$

The asymptote is given as x → -∞, f(x) = $3^{x + 2}$ → 0

∴ The horizontal asymptote is f(x) = y = 0

The y-intercept is given when x = 0, we get;

f(x) = $3^{0 + 2}$ = 9

The y-intercept is f(x) = (0, 9)

b) f(x) = $5^{1 - x}$

The asymptote is fx) = 0 as x → ∞

The asymptote is y = 0

Similar to question (1) above, the y-intercept is f(x) = $5^{1 - 0}$ = 5

The y-intercept is (0, 5)

c) f(x) = 3ˣ + 3

The asymptote is 3ˣ → 0 and f(x) → 3 as x → ∞

The asymptote is y = 3

The y-intercept is f(x) = 3⁰ + 3= 4

The y-intercept is (0, 4)

d) f(x) = 6⁻ˣ + 3

The asymptote is 6⁻ˣ → 0 and f(x) → 3 as x → ∞

The horizontal asymptote is y = 3

The y-intercept is f(x) = 6⁻⁰ + 3 = 4

The y-intercept is (0, 4)

e) f(x) = $2^{x + 3}$ - 1

The asymptote is $2^{x + 3}$ → 0 and f(x) → -1 as x → -∞

The horizontal asymptote is y = -1

The y-intercept is f(x) = $2^{0 + 3}$ - 1 = 7

The y-intercept is (0, 7)

When f(x) = 0, $2^{x + 3}$ - 1 = 0

$2^{x + 3}$ = 1

x + 3 = 0, x = -3

The x-intercept is (-3, 0)

f) $f(x) = \left (\dfrac{1}{2} \right)^{x - 2} + 2$

The asymptote is $\left (\dfrac{1}{2} \right)^{x - 2}$ → 0 and f(x) → 2 as x → ∞

The asymptote is y = 2

The y-intercept is f(x) = $f(0) = \left (\dfrac{1}{2} \right)^{0 - 2} + 2 = 6$

The y-intercept is (0, 6)

7 0

2 years ago

A shipment of 50,000 transistors arrives at a manufacturing plant. The quality control engineer at the plant obtains a random sa

Aleks04 [339]

Step-by-step explanation:

remember, the number of possible combinations to pick m out of n elements is C(n, m) = n!/(m! × (n-m)!)

50,000 transistors.

4% are defective, that means 4/100 = 1/25 of the whole.

so, the probability for one picked transistor to be defective is 1/25.

and the probability for it to work properly is then 1-1/25 = 24/25.

now, 500 picks are done.

to accept the shipment, 9 or less of these 500 picks must be defective.

the probability is then the sum of the probabilities to get

0 defective = (24/25)⁵⁰⁰

1 defective = (24/25)⁴⁹⁹×1/25 × C(500, 1)

= 24⁴⁹⁹/25⁵⁰⁰ × 500

2 defective = (24/25)⁴⁹⁸×1/25² × C(500, 2)

= 24⁴⁹⁸/25⁵⁰⁰ × 250×499

3 defective = 24⁴⁹⁷/25⁵⁰⁰ × C(500, 3) =

= 24⁴⁹⁷/25⁵⁰⁰ × 250×499×166

...

9 defective = 24⁴⁹¹/25⁵⁰⁰ × C(500, 9) =

= 24⁴⁹¹/25⁵⁰⁰ × 500×499×498×497×496×495×494×493×492×491 /

9×8×7×6×5×4×3×2 =

= 24⁴⁹¹/25⁵⁰⁰ × 50×499×166×71×31×55×494×493×41×491

best to use Excel or another form of spreadsheet to calculate all this and add it all up :

the probability that the engineer will accept the shipment is

0.004376634...

which makes sense, when you think about it, because 10 defect units in the 500 is only 2%. and since the whole shipment contains 4% defect units, it is highly unlikely that the random sample of 500 will pick so overwhelmingly the good pieces.

is the acceptance policy good ?

that completely depends on the circumstances.

what was the requirement about max. faulty rate in the first place ? if it was 2%, then the engineer's approach is basically sound.

it then further depends what are the costs resulting from a faulty unit ? that depends again on when the defect is usually found (still in manufacturing, or already out there at the customer site, or somewhere in between) and how critical the product containing such transistors is. e.g. recalls for products are extremely costly, while simply sorting the bad transistors out during the manufacturing process can be rather cheap. if there is a reliable and quick process to do so.

so, depending on repair, outage and even penalty costs it might be even advisable to have a harder limit during the sample test.

in other words - it depends on experience and the found distribution/probability curve, standard deviation, costs involved and other factors to define the best criteria for the sample test.

3 0

2 years ago

6.2? + 773 + 8.70 - 9<br> What is the degree of the polynomial?

zysi [14]

Ridjdjdntjfjfjfnfirjfbxjdidjdjxududuhdjxhxudufbxjdudjdhd

3 0

3 years ago

Plz Help! 30 Points!

Ira Lisetskai [31]

45.2 is the constant hope this helps! please give branliest

4 0

3 years ago

Read 2 more answers