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3 years ago

Which name accurately describes the figure shown below and why?

Mathematics

1 answer:

Gala2k [10]3 years ago

6 0

Option C:

The name of the figure is rhombus, because it has four sides of equal length and opposite sides that are parallel.

Solution:

In the given figure, all sides are equal and they are parallel to opposite side.

To find which name is suitable for the figure.

Trapezoid:

Trapezoid have two non-parallel sides and two parallel sides.
Four sides of trapezoid are not equal in length.

Therefore option A and option B are not true.

Rhombus:

Rhombus have four equal sides and opposite sides are parallel.
But all angles are not right angles.

Therefore Option D is not true.

Option C is the correct answer.

The name of the figure is rhombus, because it has four sides of equal length and opposite sides that are parallel.

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Y_Kistochka [10]

Answer:

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Step-by-step explanation:

"The sum of a number and six" would mean the number plus 6: n+6 and "is seven times the number" indicates that it would be equal to the product of 7 for n: 7n. According to this, the answer is that the translation of the sentence into an equation is: n+6=7n.

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3 years ago

X + 7у = 23 3х + 14у = 48

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Answer:

x + 7у = 23 3х + 14у = 48

Step-by-step explanation:

x + 7у = 23 3х + 14у = 48

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2 years ago

Samuel took out a personal loan of $13,000 at 7% simple interest. a. How much interest will he pay after 4 years? Show your work

Scorpion4ik [409]

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3 years ago

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3 years ago

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%

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.

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

Mean of X, $\mu$ = $n \times p$ = $150 \times 0.10$ = 15

Standard deviation of X, $\sigma$ = $\sqrt{np(1-p)}$ = $\sqrt{150 \times 0.10 \times (1-0.10)}$ = 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.

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3 years ago