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3 years ago

Which one is bigger in 4:7 and 5:8?

Mathematics

2 answers:

seraphim [82]3 years ago

6 0

Answer:

5/8

Step-by-step explanation:

4/7 =0.5714285714

5/8=0.625

hope it helps mark me as the brainliest

dangina [55]3 years ago

4 0

Answer:

5/8 > 4/7

Step-by-step explanation:

4/7 = 0.5714

5/8 = 0.625

5/8 > 4/7

You might be interested in

Complete the pairs of corresponding parts if △ABD ≅ △EFC.

vovangra [49]

Answer:

We conclude that:

∠A = ∠ FEC

Step-by-step explanation:

Given

△ABD ≅ △EFC

To determine:

∠A =

△ABD ≅ △EFC

So the triangles △ABD and △EFC are congruent to each other.

We know that congruent triangles have equal corresponding parts.

Please check the attached graph.

From the graph, it is clear that ∠A is correspondent to ∠E.

∠A = ∠E

From the attached figure, it is clear that:

∠F can also be denoted by ∠ FEC

∠A = ∠E

∠A = ∠ FEC

Therefore, we conclude that:

∠A = ∠ FEC

6 0

2 years ago

using the distance formula d=√(x2- x1)^2 +(y2-y^1, what is the distance between point (-3,1) and point (-2,4) rounded to the nea

defon

Answer:

$d = \sqrt{(x2 - x1) + (y2 - y1)} \\ d = \sqrt{ (- 3 + 2) + (1 + - 4)} \\ d = \sqrt{ - 1 + - 3 } \\ d = \sqrt{ - 4 } \\ d = - 2$

Step-by-step explanation:

<h2>hope it helps you<3</h2>

4 0

2 years ago

A manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective

andriy [413]

Answer:

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

(b) Expected number of defective light bulbs = 15

(c) Standard deviation of defective light bulbs = 3.67

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