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SCORPION-xisa [38]
3 years ago
8

Find the area of the shape below, giving your answer to 1 decimal place

Mathematics
2 answers:
Andre45 [30]3 years ago
7 0

Answer:the answer is 150

Step-by-step explanation:multiply 10 and 15 and you get 150

irga5000 [103]3 years ago
3 0

Answer:

189.25 \:  {cm}^{2}

Step-by-step explanation:

Area of the shape = Area of rectangle + Area of semicircle.

For rectangle:

Length = 15 cm

Width = 10 cm

For semicircle:

Diameter = 10 cm

Radius = 10/2 = 5 cm

Area of the shape

= lb +  \frac{1}{2} \pi {r}^{2}  \\  \\ = 15 \times 10 +  \frac{1}{2}   \times 3.14 \times  {(5)}^{2}  \\  \\ = 150 +  \frac{1}{2}   \times 3.14 \times 25 \\  \\  = 150 + 39.25 \\  \\  = 189.25 \:  {cm}^{2}

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What is the common difference for this arithmetic sequence?
OleMash [197]

Answer:

Concept: Series

  1. Start by taking the first and second An and finding the difference
  2. You can then prove your result by taking any ordinary An and subtracting it by An-1
  3. Hence 45-39= 6
  4. To prove our result of 6; 39-33= 6
  5. Hence the common difference is
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5 0
2 years ago
Area of paralellogram with base 15cm and height12cm
azamat

Answer:

180 cm2

Area of a polyhedron:

15(=base)x12(=height)

=180cm2

btw parallelogram is a 2D pic how you can come out a ''base'' and ''height''? It should be called a polyhedron. Your question is weird.

If so, your ''area of parallelogram'' is 15cm2 since your question is hecking  weird.

Btw it's 15cm2 not 15

Please check ur question b4 u ask

7 0
2 years ago
Read 2 more answers
Change 84% to a fraction
Sauron [17]
21/25 first you would do 84/100 then you would simplify 
3 0
3 years ago
Ezra is training for a track race. He starts by sprinting 100 yards. He gradually increases his distance, adding 5 yards a day f
Sergeu [11.5K]
On the 21st day, he sprints 205 yards.

Is that what you meant for the problem

or do u want to know how many yards he sprinted total for the 21 days
If it is,
he sprinted a total of 3390 yards
4 0
3 years ago
Read 2 more answers
A manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective
Inessa05 [86]

Answer:

(a) P(X \leq 20) = 0.9319

(b) Expected number of defective light bulbs = 15

Step-by-step explanation:

We are given that a manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective independently. A box of 150 bulbs is selected at random.

Firstly, the above situation can be represented through binomial distribution, i.e.;

P(X=r) = \binom{n}{r} p^{r} (1-p)^{2} ;x=0,1,2,3,....

where, n = number of samples taken = 150

            r = number of success

           p = probability of success which in our question is % of bulbs that

                  are defective, i.e. 10%

<em>Now, we can't calculate the required probability using binomial distribution because here n is very large(n > 30), so we will convert this distribution into normal distribution using continuity correction.</em>

So, Let X = No. of defective bulbs in a box

<u>Mean of X</u>, \mu = n \times p = 150 \times 0.10 = 15

<u>Standard deviation of X</u>, \sigma = \sqrt{np(1-p)} = \sqrt{150 \times 0.10 \times (1-0.10)} = 3.7

So, X ~ N(\mu = 15, \sigma^{2} = 3.7^{2})

Now, the z score probability distribution is given by;

                Z = \frac{X-\mu}{\sigma} ~ N(0,1)

(a) Probability that this box will contain at most 20 defective light bulbs is given by = P(X \leq 20) = P(X < 20.5)  ---- using continuity correction

    P(X < 20.5) = P( \frac{X-\mu}{\sigma} < \frac{20.5-15}{3.7} ) = P(Z < 1.49) = 0.9319

(b) Expected number of defective light bulbs found in such boxes, on average is given by = E(X) = n \times p = 150 \times 0.10 = 15.

                                           

5 0
3 years ago
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