HELP NEEDED, DESPERATE! MATHS!!! WILL AWARD BRAINLIEST!

kirill115 [55]

You just multiply g(x) and f(x) together and it gives you C :)

3 0

3 years ago

A contractor purchases a shipment of 100 transistors. It is his policy to test 10 of these transistors and to keep the shipment

Serjik [45]

Answer:

Hence the probability of the at least 9 of 10 in working condition is 0.3630492

Step-by-step explanation:

Given:

total transistors=100

defective=20

To Find:

P(X≥9)=P(X=9)+P(X=10)

Solution:

There are 20 defective and 80 working transistors.

Probability of at least 9 of 10 should be working out 80 working transistors

is given by,

P(X≥9)=P(X=9)+P(X=10)

{80C9 gives set of working transistor and 20C1 gives 20 defective transistor and 100C10 is combination of shipment of 10 transistors}

P(X≥9)= ${80C9*20C(10-9)}/(100C10)+{80C10*20(10-10)}/(100C10)$

(Use the permutation and combination calculator)

P(X≥9)=(231900297200*20/17310309456440)

+(1646492110120/17310309456440)

P(X≥9)=0.267933+0.0951162

P(X≥9)=0.3630492

4 0

3 years ago

James has just sold his motorbike for £1,890. He bought the motorbike for £2,700. What was his percentage loss? %

Usimov [2.4K]

£2.700 - £1890 = £810

£ (money) -------------- percentage (%)
2.700 -------------------- 100
810 ----------------------- x

2700*x = 810*100
2700x = 81000
$x = \frac{81000}{2700}$
$\boxed{x = 30}$

Answer:
Percentage loss = 30%

6 0

3 years ago

We have n = 100 many random variables Xi ’s, where the Xi ’s are independent and identically distributed Bernoulli random variab

777dan777 [17]

Answer:

(a) The distribution of $X=\sum\limits^{n}_{i=1}{X_{i}}$ is a Binomial distribution.

(b) The sampling distribution of the sample mean will be approximately normal.

(c) The value of $P(\bar X>0.50)$ is 0.50.

Step-by-step explanation:

It is provided that random variables $X_{i}$ are independent and identically distributed Bernoulli random variables with p = 0.50.

The random sample selected is of size, n = 100.

(a)

Theorem:

Let $X_{1},\ X_{2},\ X_{3},...\ X_{n}$ be independent Bernoulli random variables, each with parameter p, then the sum of of thee random variables, $X=X_{1}+X_{2}+X_{3}...+X_{n}$ is a Binomial random variable with parameter n and p.

Thus, the distribution of $X=\sum\limits^{n}_{i=1}{X_{i}}$ is a Binomial distribution.

(b)

According to the Central Limit Theorem if we have an unknown population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

The sample size is large, i.e. n = 100 > 30.

So, the sampling distribution of the sample mean will be approximately normal.

The mean of the distribution of sample mean is given by,

$\mu_{\bar x}=\mu=p=0.50$