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4 years ago

What number is 10% of 56

Mathematics

2 answers:

Harlamova29_29 [7]4 years ago

5 0

5.6 should be the answer

melamori03 [73]4 years ago

3 0

If you multiply 56 by 0.10 you will see that 5.6 is 10% of 56

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Mrs. Clark wants to plant rose bushes along the front of her house. The length she wants to cover is 13 feet, and she wants to a

larisa86 [58]

Answer:

Mrs Clark spends in total $119 on rose bushes.

Step-by-step explanation:

Here, the total length that needs to be covered = 13 ft

The distance between 2 consecutive rose pant = 2 ft

Cost of planting each bush = $17

Now, if we plant first plant at 1 feet,

Then the second bush will be planted at a distance of 2 ft + 1 ft = 3 ft

Third bush = Second Bush + 2 ft = 3 ft + 2 ft = At 5 ft

Fourth bush = Third Bush + 2 ft = 5 ft + 2 ft = At 7 ft

Fifth bush = Fourth Bush + 2 ft = 7 ft + 2 ft = At 9 ft

Sixth bush = Fifth Bush + 2 ft = 9 ft + 2 ft = At 11 ft

And, lastly Seventh bush = Sixth Bush + 2 ft = 11 ft + 2 ft = At 13 ft

So, as to cover a total of 13 ft, 7 bushes are planted in total which are 2 ft apart from one another.

Now, cost of planing 7 rose bushes = 7 x ( Cost of planting 1 rose bush)

= 7 x ($17) = $119

Hence, Mrs Clark spends in total $119 on rose bushes.

5 0

3 years ago

ILL GIVE BRAINLIEST

avanturin [10]

I believe it would be D sorry if I’m wrong good luck!!

8 0

3 years ago

Help, can’t figure it out, need help fast

olga55 [171]

B to the second means b * b.
The short hand for b*b is b^2 on this editor or $b^{2}$
c to the fourth means c * c * c *c
the short hand for c*c*c*c is c^4 on this editor or $c^{4}$

The answer you want is b^2/c^4 in the answer box.
If this is wrong and you have choices, please list them and I will edit or put the answer in a comment.

5 0

3 years ago

there are 8 pieces of fruit in a basket including 4 apples what is the probability that a randomly selected piece of fruit will

Helen [10]

Answer:

$\frac{1}{2}$ is the probability for picking fruits and it will be apple.

Explanation:

probability of picking apple: $\boxed{\frac{total -apples}{total-fruits} }$ → $\frac{4}{8}$ → $\frac{1}{2}$

7 0

3 years ago

Hdc produces microcomputer hard drives at four different production facilities (f1, f2, f3, and f4) hard drive production at f1,

OLEGan [10]

Let $D$ denote the event that an HD is defective, and $F_i$ the event that a particular HD was produced at facility $i$ .

You are asked to compute

$\mathbb P(F_2\mid D)$
$\mathbb P(F_4\mid D)$
$\mathbb P(D^C)$

From the definition of conditional probabilities, the first two will require that you first find $\mathbb P(D)$ . Once you have this, part (c) is trivial.

I'll demonstrate the computation for part (a). Part (b) is nearly identical.

(a)
$\mathbb P(F_2\mid D)=\dfrac{\mathbb P(F_2\cap D)}{\mathbb P(D)}$

Presumably, the facility responsible for producing a given HD is independent of whether the HD is defective or not, so $\mathbb P(F_2\cap D)=\mathbb P(F_2)\mathbb P(D)=0.20\times0.015$ .

Use the law of total probability to determine the value of the denominator:

$\mathbb P(D)=\mathbb P(D\mid F_1)+\mathbb P(D\mid F_2)+\mathbb P(D\mid F_3)+\mathbb P(D\mid F_4)$

We know each of the component probabilities because they are given explicitly: 0.015, 0.02, 0.01, and 0.03, respectively. So

$\mathbb P(D)=0.015+0.02+0.01+0.03=0.075$

and thus

$\mathbb P(F_2\mid D)=\dfrac{0.2\times0.015}{0.075}=0.04$

(b) Similarly,
$\mathbb P(F_4\mid D)=\dfrac{0.4\times0.03}{0.075}=0.16$

(c)
$\mathbb P(D^C)=1-\mathbb P(D)=0.925$

6 0

4 years ago