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disa [49]
3 years ago
7

Which of the following best describe perpendicular lines? A.lines that are not in the same plane B. Lines that meet at a 90* ang

le C. Lines that meet at a 45' angle D. Lines that are coplanar and do not intersect
Mathematics
1 answer:
frosja888 [35]3 years ago
7 0

A line is said to be perpendicular to another line if the two lines intersect at a right angle. An angle is said to be a right angle if its measure is 90 degrees.

The question is asking us to select the option which best describes a perpendicular lines.Among all the options given option B states :Lines that meet at a 90* angle.

Option B is the right answer.

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The area of a particular rectangle is 72. If the length of the rectangle is twice
soldier1979 [14.2K]

Answer:

6

Step-by-step explanation:

Suppose the width is "a".  Then the length is 2a.

2(a²) =72 Divide both sides by 2

    a²= 36

    a=6

   

4 0
3 years ago
Felipe borrowed $8000 at a rate of 16.5%, compounded monthly. Assuming he makes no payments, how much will he owe after 7 years?
Gelneren [198K]

Answer:

$25,193.17

Explanation:

Given:

• Principal Felipe borrowed, P=$8000

,

• Annual Interest Rate, r=16.5%=0.165

,

• Compounding Period, k=12 (Monthly)

,

• Time, t=7 years

We want to determine how much he will owe after 7 years.

In order to carry out this calculation, use the compound interest formula below:

A(t)=P\mleft(1+\frac{r}{k}\mright)^{tk}

Substitute the values defined above:

A(t)=8000\mleft(1+\frac{0.165}{12}\mright)^{12\times7}

Finally, simplify and round to the nearest cent.

\begin{gathered} A(t)=8000(1+0.01375)^{84} \\ =8000(1.01375)^{84} \\ =\$25,193.17 \end{gathered}

After 7 years, Felipe will owe $25,193.17.

3 0
1 year ago
A falling object accerlates from -10.0m/s to -30.0m/s how much time does that take
Leya [2.2K]

Answer:

2.04 seconds

Step-by-step explanation:

Falling objects near the surface of the earth have an acceleration of -9.81 m/s².

Acceleration is the change in velocity over change in time:

a = (v − v₀) / t

-9.81 = (-30.0 − (-10.0)) / t

-9.81 = -20.0 / t

t = 2.04

It takes 2.04 seconds.

5 0
3 years ago
Could someone me please​
mariarad [96]

Answer:

I think is 11/90.

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
Big chickens: The weights of broilers (commercially raised chickens) are approximately normally distributed with mean 1387 grams
Nataliya [291]

Answer:

a) 0.2318

b) 0.2609

c) No it is not unusual for a broiler to weigh more than 1610 grams

Step-by-step explanation:

We solve using z score formula

z-score is is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.

Mean 1387 grams and standard deviation 192 grams. Use the TI-84 Plus calculator to answer the following.

(a) What proportion of broilers weigh between 1150 and 1308 grams?

For 1150 grams

z = 1150 - 1387/192

= -1.23438

Probability value from Z-Table:

P(x = 1150) = 0.10853

For 1308 grams

z = 1308 - 1387/192

= -0.41146

Probability value from Z-Table:

P(x = 1308) = 0.34037

Proportion of broilers weigh between 1150 and 1308 grams is:

P(x = 1308) - P(x = 1150)

0.34037 - 0.10853

= 0.23184

≈ 0.2318

(b) What is the probability that a randomly selected broiler weighs more than 1510 grams?

1510 - 1387/192

= 0.64063

Probabilty value from Z-Table:

P(x<1510) = 0.73912

P(x>1510) = 1 - P(x<1510) = 0.26088

≈ 0.2609

(c) Is it unusual for a broiler to weigh more than 1610 grams?

1610- 1387/192

= 1.16146

Probability value from Z-Table:

P(x<1610) = 0.87727

P(x>1610) = 1 - P(x<1610) = 0.12273

≈ 0.1227

No it is not unusual for a broiler to weigh more than 1610 grams

8 0
3 years ago
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