All categories

3 years ago

160 bananas;20%decrease

Mathematics

2 answers:

Jobisdone [24]3 years ago

7 0

Answer:

128 banana's... I would work it all out for u but the teacher just came in my room and had a go

Natalka [10]3 years ago

6 0

Answer: 128

Step-by-step explanation:

Percent (%)' means 'out of one hundred':

p% = p 'out of one hundred',

p% is read p 'percent',

p% = p/100 = p ÷ 100.

20% = 20/100 = 20 ÷ 100 = 0.2.

100% = 100/100 = 100 ÷ 100 = 1.

You might be interested in

-x+1=0 please let me know sovle for x

Vlada [557]

Subtract 1 from +1 and 0 and then divide the negative from x and -1 which gives u with x=1

6 0

3 years ago

Read 2 more answers

Find an equation of each line given one point and the slope : (4, 8); slope 8

Fudgin [204]

Step-by-step explanation:

in such a case the easiest approach might be using the point-slope form of a line equation :

y - y1 = m(x - x1)

with m being the slope, and (x1, y1) being a point on the line.

so, in our case that would be

y - 8 = 8(x - 4)

simplified this would be

y - 8 = 8x - 32

y = 8x - 24

FYI - this would be then the slope-intercept form (-24 would be the intercept point on the y-axis).

3 0

2 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.

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,

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