All categories

3 years ago

What is an dividend in a math equation?

Mathematics

1 answer:

Harlamova29_29 [7]3 years ago

4 0

Example:

35 ÷ 5 = 7

35 = dividend
5 = divisor
7 = quotient

You might be interested in

Michaela works in a bank she is paid £19200 per annum how much does she get per month

Step2247 [10]

in a year she earns=19200

So in a month she earns=19200/12

=1600

So she earns 1600 every month.

If the answer is correct then make me brain list.please

4 0

3 years ago

List the < from the largest to smallest for ABC

devlian [24]

Answer: B

Step-by-step explanation:

Since $AB < AC < BC$ , it follows that $\angle C < \angle B < \angle A$

4 0

2 years ago

Select the correct answer.

tankabanditka [31]

Answer:

Saving (Start) = 1000

Add to savings = 300

y = 300x + 1000

Motorcycle = 5000

Payment = 250

y = 5000 - 250x

Answer is C

Step-by-step explanation:

8 0

3 years ago

Find two solutions to the equation (x^3 − 64)(x^5 − 1) = 0.

Rashid [163]

Answer:

the two roots are x = 1 and x = 4

Step-by-step explanation:

Data provided in the question:

(x³ − 64) (x⁵ − 1) = 0.

Now,

for the above relation to be true the following condition must be followed:

Either (x³ − 64) = 0 ............(1)

(x⁵ − 1) = 0 ..........(2)

Therefore,

considering the first equation, we have

(x³ − 64) = 0

adding 64 both sides, we get

x³ − 64 + 64 = 0 + 64

x³ = 64

taking the cube root both the sides, we have

∛x³ = ∛64

x = ∛(4 × 4 × 4)

x = 4

similarly considering the equation (2) , we have

(x⁵ − 1) = 0

adding the number 1 both the sides, we get

x⁵ − 1 + 1 = 0 + 1

x⁵ = 1

taking the fifth root both the sides, we get

$\sqrt[5]{x^5}=\sqrt[5]{1}$

also,

1 can be written as 1⁵

therefore,

$\sqrt[5]{x^5}=\sqrt[5]{1^5}$

x = 1

Hence,

the two roots are x = 1 and x = 4

4 0

2 years ago

A quality-conscious disk manufacturer wishes to know the fraction of disks his company makes which are defective. Step 2 of 2 :

const2013 [10]

Answer:

The 80% confidence interval for the population proportion of disks which are defective is (0.059, 0.079).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

Suppose a sample of 1067 floppy disks is drawn. Of these disks, 74 were defective.

This means that $n = 1067, \pi = \frac{74}{1067} = 0.069$

80% confidence level

So $\alpha = 0.2$ , z is the value of Z that has a pvalue of $1 - \frac{0.2}{2} = 0.9$ , so $Z = 1.28$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.069 - 1.28\sqrt{\frac{0.069*0.931}{1067}} = 0.059$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.069 + 1.28\sqrt{\frac{0.069*0.931}{1067}} = 0.079$

The 80% confidence interval for the population proportion of disks which are defective is (0.059, 0.079).

7 0

2 years ago