Factorise completely 4ab+6ac-6bd-9cd

Zen spent $255 on a bag and a belt. She wanted to buy another

Galina-37 [17]

Answer:

A)$ 45

B) $105

Step-by-step explanation:

Bag and a belt cost $255

Let bag = x

Let belt = y

X+y= 255 equation 1

Let total money be z first

Remaining money= z-255

X-30 = z-255

Y +15 = z-255

Equating the left side of the equation

X+30 = y+15

X-y= 45 equation 2

Solving simultaneously

X+y= 255

X-y= 45

2x = 300

X= 150

If x= 150

150-y= 45

150-45= y

105=y

Bag = $150

Belt = $105

Bag Is 150-105 more than the belt

150-105= $45

8 0

3 years ago

The product of odd numbers is always even true or false ?

likoan [24]

Answer:

false

Step-by-step explanation:

let's say we have two odd numbers

2k+1 and 2t+1

$(2k+1)(2t+1)\\\\=4kt+2k+2t+1\\\\=2(2kt+k+t)+1$

so we see it's odd

6 0

3 years ago

Read 2 more answers

A population has a mean of 200 and a standard deviation of 50. Suppose a sample of size 100 is selected and x is used to estimat

zmey [24]

Answer:

a) 0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b) 0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .