Ask question

Ask question

All categories

3 years ago

6

Reuben made 4 identical necklaces, each having beads and a pendant. The total cost of the beads and pendants for all 4 necklaces

was $24.40. If the beads cost $2.20 for each necklace, how much did each pendant cost?

2 answers:

zysi [14]3 years ago

5 0

Answer:

$3.90

Step-by-step explanation:

The total cost was $24.40.

We know the cost for beads for each necklace is also $2.20, multiply this by four to find the cost of the beads for all the necklaces.

$2.20 × 4 = $8.80

Now we will subtract this from the total price.

$24.40 - $8.80 = $15.60

$15.60 Is the remaining cost for the pendants of all four necklaces. We will now divide this by four to find the price of a single pendant.

$15.60 ÷ 4 = $3.90

<u>This means the cost of each pendant is $3.90</u>

Oksi-84 [34.3K]3 years ago

4 0

2.20•4=8.80 so if you subtract that by 24.40 you get 15.60 and if you divide that by 4 you get 3.90 for each pendant. If you wanna check it you can add 3.90 and 2.20 together and multiply it by 4 and you get 24.40

You might be interested in

Find the trigonometric integral. (Use C for the constant of integration.)

Olin [163]

Suppose we let $u=\sin\theta+\cos\theta$ , so that $\mathrm du=(\cos\theta-\sin\theta)\,\mathrm d\theta$ .

Also, recall the double angle identity for cosine:

$\cos(2\theta)=\cos^2\theta-\sin^2\theta=(\cos\theta-\sin\theta)(\cos\theta+\sin\theta)$

So, we can rewrite and compute the integral using the substitution, as

$\displaystyle\int\cos(2\theta)(\sin\theta+\cos\theta)^3\,\mathrm d\theta$

$=\displaystyle\int u\cdot u^3\,\mathrm du$

$=\displaystyle\int u^4\,\mathrm du$

$=\dfrac{u^5}5+C$

$=\boxed{\dfrac{(\cos\theta+\sin\theta)^5}5+C}$

4 0

3 years ago

SURFACE AREAS AND VOLUMES (10Th Grade)

Mekhanik [1.2K]

Answer: here is ur answer

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

The state medical school has discovered a new test for tuberculosis. (If the test indicates a person has tuberculosis, the

Inessa [10]

The probability according to the given scenario is:

(a) <em>0.0395</em>

(b) <em>0.95</em>

(c) <em>0.076</em>

The given values are:

$P(B/A) = 0.79$
$P(A) = 0.05$

(a)

The probability that the b/w has tuberculosis as well as having (+) test will be:

→ $P(A \cap B) = P(B/A).P(A)$

By substituting the values, we get

→ $= 0.79\times 0.05$

→ $= 0.0395$

(b)

The probability that a person does not have tuberculosis will be:

→ $P(A') = 1-P(A)$

→ $=1-0.05$

→ $= 0.95$

(c)

The person does not have tuberculosis as well as have a (+) test, the probability will be:

→ $P(A' \cap B) = P(A').P(B/A')$

→ $= 0.95\times 0.08$

→ $= 0.076$

Learn more:

brainly.com/question/14816227

8 0

2 years ago

Please help please please

Grace [21]

$\Omega=\{1;\ 2;\ 3;\ 4;\ 5;\ 6;\ 7;\ 8;\ 9;\ 10\}\\\\|\Omega|=10\\\\A=\{1;\ 3;\ 6;\ 7\};\ B=\{2;\ 3\}\\\\A\ \cup\ B=\{1;\ 2;\ 3;\ 6;\ 7\}\\\\|A\ \cup\ B|=5\\\\P(A\ \cup\ B)=\dfrac{|A\ \cup\ B|}{|\Omega|}\to P(A\ \cup\ B)=\dfrac{5}{10}=\dfrac{1}{2}=0.5=0.50$

4 0

3 years ago

Read 2 more answers

if sally has 5 apples and 4 oranges and oranges weigh 5 lbs per orange and apples weigh 20 lbs an apple how much is sallys final

kotykmax [81]

4 oranges multiplied by the 5 pounds =20 + the other 20 = 40. Your answer is 40

5 0

3 years ago

Read 2 more answers