All categories

3 years ago

A jewellery shop sells 240 necklaces in a month.

1 answer:

8 0

3:1 is the ratio of online sales to shop sales.

What I did:
240-180=60
180/60=3
Therefore the ratio would be 3:1

Hope this helped:)

You might be interested in

Which of the following is an extraneous solution of sqrt 4x + 41 = x+5

stiv31 [10]

Answer:

x=-8

Step-by-step explanation:

$\sqrt{4x+41}=x+5$

on squaring both sides

$4x+41=(x+5)^2\\\\4x+41=x^2+10x+25\\\\x^2+6x-16=0\\\\x^2+8x-2x-16=0\\\\x(x+8)-2(x+8)=0\\\\(x-2)(x+8)=0$

either x-2=0 or x+8=0

either x=2 or x=-8

Putting x=2 in original equation, we get

$\sqrt{4\times 2+41}=2+5$

7=7 hence, it is not an extraneous solution

Putting x=-8 in original equation

$\sqrt{4\times (-8)+41}=-8+5$

i.e. 3=-3

Hence, x=-8 is an extraneous solution (since it doesn't satisfy the original equation)

3 0

3 years ago

Read 2 more answers

Find the sum of the positive integers less than 200 which are not multiples of 4 and 7

taurus [48]

Answer:

$12942$ is the sum of positive integers between $1$ (inclusive) and $199$ (inclusive) that are not multiples of $4$ and not multiples $7$ .

Step-by-step explanation:

For an arithmetic series with:

$a_1$ as the first term,
$a_n$ as the last term, and
$d$ as the common difference,

there would be $\displaystyle \left(\frac{a_n - a_1}{d} + 1\right)$ terms, where as the sum would be $\displaystyle \frac{1}{2}\, \displaystyle \underbrace{\left(\frac{a_n - a_1}{d} + 1\right)}_\text{number of terms}\, (a_1 + a_n)$ .

Positive integers between $1$ (inclusive) and $199$ (inclusive) include:

$1,\, 2,\, \dots,\, 199$ .

The common difference of this arithmetic series is $1$ . There would be $(199 - 1) + 1 = 199$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times ((199 - 1) + 1) \times (1 + 199) = 19900 \end{aligned}$ .

Similarly, positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $4$ include:

$4,\, 8,\, \dots,\, 196$ .

The common difference of this arithmetic series is $4$ . There would be $(196 - 4) / 4 + 1 = 49$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 49 \times (4 + 196) = 4900 \end{aligned}$

Positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $7$ include:

$7,\, 14,\, \dots,\, 196$ .

The common difference of this arithmetic series is $7$ . There would be $(196 - 7) / 7 + 1 = 28$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 28 \times (7 + 196) = 2842 \end{aligned}$

Positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $28$ (integers that are both multiples of $4$ and multiples of $7$ ) include:

$28,\, 56,\, \dots,\, 196$ .

The common difference of this arithmetic series is $28$ . There would be $(196 - 28) / 28 + 1 = 7$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 7 \times (28 + 196) = 784 \end{aligned}$ .

The requested sum will be equal to:

the sum of all integers from $1$ to $199$ ,
minus the sum of all integer multiples of $4$ between $1\!$ and $199\!$ , and the sum integer multiples of $7$ between $1$ and $199$ ,
plus the sum of all integer multiples of $28$ between $1$ and $199$ - these numbers were subtracted twice in the previous step and should be added back to the sum once.

That is:

$19900 - 4900 - 2842 + 784 = 12942$ .

8 0

3 years ago

Travelers who fail to cancel their hotel reservations when they have no intention of showing up are commonly referred to as no-s

notsponge [240]

Answer:

a) 0.0523 = 5.23% probability that at least two of the four selected will turn to be no-shows.

b) 0 is the most likely value for X.

Step-by-step explanation:

For each traveler who made a reservation, there are only two possible outcomes. Either they show up, or they do not. The probability of a traveler showing up is independent of other travelers. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$