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kotegsom [21]
2 years ago
10

Construct a rectangle with sides 6cm and 4cm.​

Mathematics
2 answers:
Bess [88]2 years ago
4 0

Answer:

Step-by-step explanation:

Maru [420]2 years ago
4 0

Answer:

  1. First, draw a line segment AB = 6 cm.
  2. Draw a ray AX with pencil and compass such that ∠BAX = 90°. Draw an arc of 4 cm with point A which intersects AX at D.
  3. Similarly, draw ray BY such that ∠ABY = 90°. Draw an arc of 4 cm with point B which intersects BY at C.
  4. Join CD.
  5. Thus ABCD is the required rectangle.

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A bag contains 8 purple marbles & 12 green marbles. You choose one marble out of the bag without looking.
MrRissso [65]

Answer:

1. P(P) = 8/20 = 0.4

2. P(G) = 12/20 = 0.6

Step-by-step explanation:

Given;

Number of green marbles G = 12

Number of purple marbles P = 8

Total T = 12+8 = 20

The probability that you choose a purple marble P(P) is;

P(P) = number of purple marbles/total number of marbles

P(P) = P/T = 8/20 = 0.4

P(P) = 0.4

The probability that you choose a Green marble P(G) is;

P(G) = number of Green marbles/total number of marbles

P(G) = G/T = 12/20 = 0.6

P(G) = 0.6

6 0
3 years ago
HELP ASAP PLEASE!!!!
DENIUS [597]

Answer:

q=(1,5) t=(-2,3)r=(3,-1)s=(0,0)

Step-by-step explanation:

6 0
2 years ago
Emma has 7 bottles: 3 bottles that contain juice, 2 bottles
UNO [17]

Answer: Out of 7 bottles, 2 bottles contain soda. This means that 2 bottles= 29% out of 100%. 7/2 (7 divided by 2) = 0.285. 0.285 can be rounded up to 0.29, meaning that the approximate percentage of the bottles that contain soda is 29%.

4 0
3 years ago
which shows the students in order from greatest test score to lease test score Alex 0.95 Octavia 16/20 Tonya 9/10 Wilson 0.87 ​
Nonamiya [84]

Answer:

Alex (0.95), Tonya (0.90), Wilson (0.87), Octavia (0.80)

Step-by-step explanation:

To solve this, we have to convert all of the scores into either a decimal or a fraction. In this case, I will be solving all of them into decimals since it is easier.

Alex: 0.95

Octavia: 16/20 = 0.80

Tonya: 9/10 = 0.90

Wilson: 0.87

Therefore, we can order these from greatest score to least score.

Alex (0.95), Tonya (0.90), Wilson (0.87), Octavia (0.80)

4 0
2 years ago
The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3
In-s [12.5K]

Answer:

a) There is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

c) There is a 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3 minutes. This means that \mu = 8.3, \sigma = 3.3.

(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?

We are working with a sample mean of 37 jets. So we have that:

s = \frac{3.3}{\sqrt{37}} = 0.5425

Total time of 320 minutes for 37 jets, so

X = \frac{320}{37} = 8.65

This probability is the pvalue of Z when X = 8.65. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{8.65 - 8.3}{0.5425}

Z = 0.65

Z = 0.65 has a pvalue of 0.7422. This means that there is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?

Total time of 275 minutes for 37 jets, so

X = \frac{275}{37} = 7.43

This probability is subtracted by the pvalue of Z when X = 7.43

Z = \frac{X - \mu}{\sigma}

Z = \frac{7.43 - 8.3}{0.5425}

Z = -1.60

Z = -1.60 has a pvalue of 0.0548.

There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?

Total time of 320 minutes for 37 jets, so

X = \frac{320}{37} = 8.65

Total time of 275 minutes for 37 jets, so

X = \frac{275}{37} = 7.43

This probability is the pvalue of Z when X = 8.65 subtracted by the pvalue of Z when X = 7.43.

So:

From a), we have that for X = 8.65, we have Z = 0.65, that has a pvalue of 0.7422.

From b), we have that for X = 7.43, we have Z = -1.60, that has a pvalue of 0.0548.

So there is a 0.7422 - 0.0548 = 0.6874 = 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.

7 0
3 years ago
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