Ask question

Ask question

All categories

chubhunter [2.5K]

2 years ago

14

If the area of a rhombus is 120, and d1 is 4 cm more than twice the length of the shorter one, d2. What are the lengths of the d

iagonals of the rhombus? Round to three decimals

1 answer:

melomori [17]2 years ago

5 0

The formula for the diagonal of a rhombus is derived using the area of the rhombus. In other words, if the area and one of the diagonals is given, then the other diagonal can be calculated using the formula, p = (2 × Area)/q, where 'p' and 'q' are the two diagonals of the rhombus.

You might be interested in

Find the distance between the points. (5,10) (5,4)

german

Answer:

.

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Which expression is equivalent to (0.5n−0.3)−(0.8n−0.9)?

Bezzdna [24]

Answer:

-0.3n + 0.6

Step-by-step explanation:

if we write out the entire expression we have:

0.5n - 0.3 - 0.8n + 0.9 (two negative signs create a positive sign)

then we add and subtract

and receive

-0.3n + 0.6

7 0

3 years ago

Read 2 more answers

What is the standard deviation of the following data set rounded to the nearest tenth? 56 ,78 ,123 , 34, 67, 91, 20

quester [9]

Answer:

The standard deviation of the following data set is 32.2

Step-by-step explanation:

step 1

Find the mean

we have

$[56,78,123,34,67,9,20]$

Sum the data and divided by the number of elements

$[56+78+123+34+67+91+20]/7=469/7=67$

step 2

For each number: subtract the Mean and square the result

$[(56-67)^{2},(78-67)^{2},(123-67)^{2},(34-67)^{2},(67-67)^{2},(91-67)^{2},(20-67)^{2}]$

$[121,121,3,136,1,089,0,576,2,209]$

step 3

Work out the mean of those squared differences

$[121+121+3,136+1,089+0+576+2,209]/7=1,036$

This value is called the "Variance"

step 4

Take the square root of the variance

$standard\ deviation=\sqrt{1,036}=32.2$

6 0

3 years ago

A trapezoid has an area of 50 square units. The lengths of the bases are 10 units and 15 units. find the height.

Lera25 [3.4K]

Step-by-step explanation:

area = ( base 1 + base 2 ) ÷ 2 x height

50 = ( 10 + 15) ÷2 x height

50 ÷ 12.5 = height

height = 4

5 0

3 years ago

This exercise illustrates that poor quality can affect schedules and costs. A manufacturing process has 90 customer orders to fi

svp [43]

Answer:

a) 0.0645 = 6.45% probability that the 90 orders can be filled without reordering components.

b) 0.4062 = 40.62% probability that the 100 orders can be filled without reordering components.

c) 0.9034 = 90.34% probability that the 100 orders can be filled without reordering components

Step-by-step explanation:

For each component, there are only two possible outcomes. Either it is defective, or it is not. The components can be assumed to be independent. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

3% of the components are identified as defective

This means that $p = 0.03$

a. If the manufacturer stocks 90 components, what is the probability that the 90 orders can be filled without reordering components?

0 defective in a set of 90, which is $P(X = 0)$ when $n = 90$ . So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{90,0}.(0.03)^{0}.(0.97)^{90} = 0.0645$

0.0645 = 6.45% probability that the 90 orders can be filled without reordering components.

b. If the manufacturer stocks 102 components, what is the probability that the 100 orders can be filled without reordering components?

At most 102 - 100 = 2 defective in a set of 102, so $P(X \leq 2)$ when $n = 102$

Then

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{102,0}.(0.03)^{0}.(0.97)^{102} = 0.0447$

$P(X = 1) = C_{102,0}.(0.03)^{1}.(0.97)^{101} = 0.1411$

$P(X = 2) = C_{102,2}.(0.03)^{2}.(0.97)^{100} = 0.2204$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0447 + 0.1411 + 0.2204 = 0.4062$

0.4062 = 40.62% probability that the 100 orders can be filled without reordering components.

c. If the manufacturer stocks 105 components, what is the probability that the 100 orders can be filled without reordering components?

At most 105 - 100 = 5 defective in a set of 105, so $P(X \leq 5)$ when $n = 105$

Then

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{105,0}.(0.03)^{0}.(0.97)^{105} = 0.0408$

$P(X = 1) = C_{105,0}.(0.03)^{1}.(0.97)^{104} = 0.1326$

$P(X = 2) = C_{105,2}.(0.03)^{2}.(0.97)^{103} = 0.2133$

$P(X = 3) = C_{105,3}.(0.03)^{3}.(0.97)^{102} = 0.2265$

$P(X = 4) = C_{105,4}.(0.03)^{4}.(0.97)^{101} = 0.1786$

$P(X = 5) = C_{105,5}.(0.03)^{5}.(0.97)^{100} = 0.1116$

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0408 + 0.1326 + 0.2133 + 0.2265 + 0.1786 + 0.1116 = 0.9034$

0.9034 = 90.34% probability that the 100 orders can be filled without reordering components

3 0

3 years ago