mr tonneman buys one 2 pound lobster and 1 pound 4 ounces of shrimp. find the ratio the weight of the lobster to the shrimp

A small dealership leased 21 Suburu Outbacks on 2-year leases. When the cars were returned at the end of the lease, the mileage

Liula [17]

Answer:

yes 30 is the answer

Step-by-step explanation:

4 0

2 years ago

For quality control purposes, a company that manufactures sim chips for cell/smart phones routinely takes samples from its pro

ololo11 [35]

Answer:

There is a 22.42% probability that a sample in this size has 2 imperfections.

Step-by-step explanation:

For each chip, there are only two possible outcomes. Either they are imperfect, or they are not.

This means that we can solve this problem using binomial distribution probability concepts.

Binomial probability distribution

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

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

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

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

And $\pi$ is the probability of X happening.

In this problem

A success is a chip being imperfect. Suppose the average number of imperfections per 1000 sim chips is 3. So $\pi = \frac{3}{1000} = 0.003$ .

What is the probability that a sample this size (1000 chips) has 2 imperfections?

The sample has 1000 chips, wo $n = 1000$

We want P(X = 2).

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

$P(X = 2) = C_{1000,2}.(0.003)^{2}.(0.997)^{998} = 0.2242$

There is a 22.42% probability that a sample in this size has 2 imperfections.

7 0

2 years ago

Use Pythagorean identities to prove whether ΔLMN is a right, acute, or obtuse triangle. Show all work for full credit.

diamong [38]

Answer: The given triangle LMN is an obtuse-angled triangle.

Step-by-step explanation: We are given to use Pythagorean identities to prove whether ΔLMN is a right, acute, or obtuse triangle.

From the figure, we note that

in ΔLMN, LM = 5 units, MN = 13 units and LN = 14 units.

We know that a triangle with sides a units, b units and c units (a > b, c) is said to be

(i) Right-angled triangle if $b^2+c^2=a^2,$

(ii) Acute-angled triangle if $b^2+c^2>a^2,$

(iii) Obtuse-angled triangle if $b^2+c^2$

For the given triangle LMN, we have

a = 14, b = 13 and c = 5.

So,

$b^2+c^2=13^2+5^2=169+25=194,\\\\a^2=14^2=196.$

Therefore, $b^2+c^2$

Thus, the given triangle LMN is an obtuse-angled triangle.

5 0

2 years ago

About how many more inches of rain did Asheville get than Wichita? About how many more days did it rain in Asheville than Wichit

zaharov [31]

Answer:

Par 1) 19.10 inches

Part 2) 39 days

Step-by-step explanation:

we know that

Asheville

47.71 in

124 days

Wichita

28.61 in

85 days

Part 1) About how many more inches of rain did Asheville get than Wichita?

In this part subtract the number of inches of rain in Wichita from the number of inches of rain in Asheville

so

$47.71-28.61=19.10\ in$

Part 2) About how many more days did it rain in Asheville than Wichita?

n this part subtract the number of days of rain in Wichita from the number of days of rain in Asheville

so

$124-85=39\ days$

4 0

2 years ago

The indicated function y1(x) is a solution of the given differential equation. Use reduction of order or formula (5) in Section

LUCKY_DIMON [66]

Answer:

y2 = C1xe^(4x)

Step-by-step explanation:

Given that y1 = e^(4x) is a solution to the differential equation

y'' - 8y' + 16y = 0

We want to find the second solution y2 of the equation using the method of reduction of order.

Let

y2 = uy1

Because y2 is a solution to the differential equation, it satisfies

y2'' - 8y2' + 16y2 = 0

y2 = ue^(4x)

y2' = u'e^(4x) + 4ue^(4x)

y2'' = u''e^(4x) + 4u'e^(4x) + 4u'e^(4x) + 16ue^(4x)

= u''e^(4x) + 8u'e^(4x) + 16ue^(4x)

Using these,

y2'' - 8y2' + 16y2 =

[u''e^(4x) + 8u'e^(4x) + 16ue^(4x)] - 8[u'e^(4x) + 4ue^(4x)] + 16ue^(4x) = 0

u''e^(4x) = 0

Let w = u', then w' = u''

w'e^(4x) = 0

w' = 0

Integrating this, we have

w = C1

But w = u'

u' = C1

Integrating again, we have

u = C1x

But y2 = ue^(4x)

y2 = C1xe^(4x)

And this is the second solution

5 0

2 years ago