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3 years ago

Someone help! It is urgent!!!

Mathematics

1 answer:

djverab [1.8K]3 years ago

7 0

Answer:

a. f(-1)=12

b. f(2t)=16t²-6t+5

c. f(t-2)=4t²-19t+27

Step-by-step explanation:

For a, b, c, we are given an input. We plug that into f(x) to find our answers.

a. f(-1)=12

f(-1)=4(-1)²-3(-1)+5

f(-1)=4+3+5

f(-1)=12

-------------------------------------------------------------------------------------------------------------

b. f(2t)=16t²-6t+5

f(2t)=4(2t)²-3(2t)+5

f(2t)=4(4t²)-6t+5

f(2t)=16t²-6t+5

-------------------------------------------------------------------------------------------------------------

c. f(t-2)=4t²-19t+27

f(t-2)=4(t-2)²-3(t-2)+5

f(t-2)=4(t²-4t+4)-3t+6+5

f(t-2)=4t²-16t+16-3t+6+5

f(t-2)=4t²-19t+27

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A manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective

andriy [413]

Answer:

(a) P(X $\leq$ 20) = 0.9319

(b) Expected number of defective light bulbs = 15

(c) Standard deviation of defective light bulbs = 3.67

Step-by-step explanation:

We are given that a manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective independently. A box of 150 bulbs is selected at random.

Firstly, the above situation can be represented through binomial distribution, i.e.;

$P(X=r) = \binom{n}{r} p^{r} (1-p)^{2} ;x=0,1,2,3,....$

where, n = number of samples taken = 150

r = number of success

p = probability of success which in our question is % of bulbs that

are defective, i.e. 10%

Now, we can't calculate the required probability using binomial distribution because here n is very large(n > 30), so we will convert this distribution into normal distribution using continuity correction.

So, Let X = No. of defective bulbs in a box

Mean of X, $\mu$ = $n \times p$ = $150 \times 0.10$ = 15

Standard deviation of X, $\sigma$ = $\sqrt{np(1-p)}$ = $\sqrt{150 \times 0.10 \times (1-0.10)}$ = 3.7

So, X ~ N( $\mu = 15, \sigma^{2} = 3.7^{2})$

Now, the z score probability distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

(a) Probability that this box will contain at most 20 defective light bulbs is given by = P(X $\leq$ 20) = P(X < 20.5) ---- using continuity correction

P(X < 20.5) = P( $\frac{X-\mu}{\sigma}$ < $\frac{20.5-15}{3.7}$ ) = P(Z < 1.49) = 0.9319

(b) Expected number of defective light bulbs found in such boxes, on average is given by = E(X) = $n \times p$ = $150 \times 0.10$ = 15.

Standard deviation of defective light bulbs is given by = S.D. = $\sqrt{np(1-p)}$ = $\sqrt{150 \times 0.10 \times (1-0.10)}$ = 3.67

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3 years ago

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liubo4ka [24]

Answer:

12/13

Step-by-step explanation:

36/39

Divide by 3

12/13

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2 years ago

nancy spent 7/8 hour working out at the gym she spent 5/7 of that time lifting weights what fraction of an hour did sje spentld

Juli2301 [7.4K]

$\frac{9}{56}\\$ Fraction of hour was spent by NAncy in lifting weights

Step-by-step explanation:

Time spent at the gym by Nancy = $\frac{7}{8}\,hour$

Time spent at lifting weights = $\frac{5}{7}\,hour$

What fraction of hour she spent in lifting weights?

Solving:

Fraction of hour she spent in lifting weights= Time spend at the gym-Time spent at lifting weights

Fraction of hour she spent in lifting weights= $\frac{7}{8}\,hour-\frac{5}{7}\,hour$

$=\frac{49}{56}-\frac{40}{56}\\=\frac{49-40}{56}\\=\frac{9}{56}\\$

So, $\frac{9}{56}\\$ Fraction of hour was spent by Nancy in lifting weights

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