8h + 16h -12 = 24h -12

Please help me it is timed and i am running outta time

Westkost [7]

Answer:

In ∆RST and ∆XYZ

RS=YZ(S)

angle RST=angle XYZ(A)

ST=YZ(S)

hence ∆RST ≠~∆XYZ

7 0

3 years ago

Need help can somebody answer this question for me

nata0808 [166]

Answer:

f ≥ 40

d ≤ 200

Step-by-step explanation:

We have to write inequalities to represent the situation given:

a) The first condition is given as the vehicle's fuel efficiency is no less than 40 miles per gallon.

That means the vehicle's fuel efficiency is greater than or equal to 40 miles per gallon.

If we represent f as the vehicle's fuel efficiency ( in miles per gallon ), we can write the inequality as

f ≥ 40 (Answer)

b) The second condition is given as the distance to the nearest exit door is at most 200 feet.

That means the distance to the nearest exit door is less than or equal to 200 feet.

Therefore, if we represent d ( in feet ) as the distance to the nearest exit door, then we can write the inequality as

d ≤ 200 (Answer)

8 0

3 years ago

Parents wish to have $130,000 available for a child's education. If the child is now 6 years old, how much money must be set asi

n200080 [17]

Answer:

Step-by-step explanation:

We know that:

FV = $130,000

r = 4% (semiannually )

We need to find number of periods (n) , the child is now 6 years old, until the child is 18 is 12 years. But it is the rate is compounded semiannually, so n = 12*2 = 24

We have the fomula to find out periodic payment (see the attached photo)

=> pmt = 3326.288

8 0

4 years ago

Plz help i need it!!!!!!!

Flauer [41]

Answer:

b

Step-by-step explanation:

please answer my last question on my profile for 23 points

8 0

3 years ago

A batch of 445 containers for frozen orange juice contains 3 that are defective. Two are selected, at random, without replacemen

Elan Coil [88]

Answer:

When Two containers are selected

(a) Probability that the second one selected is defective given that the first one was defective = 0.00450

(b) Probability that both are defective = 0.0112461

(c) Probability that both are acceptable = 0.986

2. When Three containers are selected

(a) Probability that the third one selected is defective given that the first and second one selected were defective = 0.002.

(b) Probability that the third one selected is defective given that the first one selected was defective and the second one selected was okay = 0.00451

(c) Probability that all three are defective = 6.855 x $10^{-8}$ .

Step-by-step explanation:

We are given that a batch of 445 containers for frozen orange juice contains 3 defective ones i.e.

Total containers = 445

Defective ones = 3

Non - Defective ones = 442 { Acceptable ones}

Two containers are selected, at random, without replacement from the batch.

(a) Probability that the second one selected is defective given that the first one was defective is given by;

Since we had selected one defective so for selecting second the available

containers are 444 and available defective ones are 2 because once

chosen they are not replaced.

Hence, Probability that the second one selected is defective given that the first one was defective = $\frac{2}{444}$ = 0.00450

(b) Probability that both are defective = P(first being defective) +

P(Second being defective)

= $\frac{3}{445} + \frac{2}{444}$ = 0.0112461

(c) Probability that both are acceptable = P(First acceptable) + P(Second acceptable)

Since, total number of acceptable containers are 442 and total containers are 445.

So, Required Probability = $\frac{442}{445}*\frac{441}{444}$ = 0.986

Three containers are selected, at random, without replacement from the batch.

(a) Probability that the third one selected is defective given that the first and second one selected were defective is given by;

Since we had selected two defective containers so now for selecting third defective one, the available total containers are 443 and available defective container is 1 .

Therefore, Probability that the third one selected is defective given that the first and second one selected were defective = $\frac{1}{443}$ = 0.002.

(b) Probability that the third one selected is defective given that the first one selected was defective and the second one selected was okay is given by;

Since we had selected two containers so for selecting third container to be defective, the total containers available are 443 and available defective containers are 2 as one had been selected.

Hence, Required probability = $\frac{2}{443}$ = 0.00451 .

(c) Probability that all three are defective = P(First being defective) +

P(Second being defective) + P(Third being defective)

= $\frac{3}{445}* \frac{2}{444} * \frac{1}{443}$ = 6.855 x $10^{-8}$ .

5 0

4 years ago