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4 years ago

Which statement is true

Mathematics

2 answers:

anastassius [24]4 years ago

7 0

The answer is letter c

Rudiy274 years ago

5 0

The third one should be true, only the same side of similar shapes will be similar. so in here the letters doesn't match the same side and the only on matches is C

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one batch of vegetable soup uses 2 cups of chopped onions, 1.5 cups of celery, and 1 cup of chopped carrots. select the ratio th

kicyunya [14]

So what you need to do is add all of the veggies up. 2+1.5+1= 4.5
Then you write the ratio out. 1:4.5.
This can also be written as 2/9 or .222 repeating

7 0

4 years ago

Read 2 more answers

vanessa drew a figure and said that it had an infinite number of lines of symmetry. what figure did she draw?

Korolek [52]

A circle
That's a very easy one cuz if u try u will find it to be correct

5 0

3 years ago

Read 2 more answers

An aircraft emergency locator transmitter (ELT) is a device designed to transmit a signal in the case of a crash. The ACME Manuf

iren [92.7K]

Answer:

a) P(ACME) = 0.7

b) P(ACME/D) = 0.5976

Step-by-step explanation:

Taking into account that ACME manufacturing company makes 70% of the ELTs, if a locator is randomly selected from the general population, the probability that it was made by ACME manufacturing Company is 0.7. So:

P(ACME) = 0.7

Then, the probability P(ACME/D) that a randomly selected locator was made by ACME given that the locator is defective is calculated as:

P(ACME/D) = P(ACME∩D)/P(D)

Where the probability that a locator is defective is:

P(D) = P(ACME∩D) + P(B. BUNNY∩D) + P(W. E. COYOTE∩D)

So, the probability P(ACME∩D) that a locator was made by ACME and is defective is:

P(ACME∩D) = 0.7*0.035 = 0.0245

Because 0.035 is the rate of defects in ACME

At the same way, P(B. BUNNY∩D) and P(W. E. COYOTE∩D) are equal to:

P(B. BUNNY∩D) = 0.25*0.05 = 0.0125

P(W. E. COYOTE∩D) = 0.05*0.08 = 0.004

Finally, P(D) and P(ACME/D) is equal to:

P(D) = 0.0245 + 0.0125 + 0.004 = 0.041

P(ACME/D) = 0.0245/0.041 = 0.5976

5 0

3 years ago

Assume the world catch of fish in 1950 was 12 million tons and in 1955 it was 96 million tons.

Rama09 [41]

The rate of change is 16.8 million tons per year

Solution:

Given that, the world catch of fish in 1950 was 12 million tons and in 1955 it was 96 million tons

The average rate of change is given by formula:

$\text{ Average rate of change } = \frac{\text{ change in value }}{\text{ number of years }}$

Value in 1950 = 12 million tons

Value in 1955 = 96 million tons

Change in value = value in 1955 - value in 1950

Change in value = 96 million - 12 million = 84 million tons

Number of years = 1950 to 1955 = 5 years

Substitute the given values in formula,

$\text{ Average rate of change } = \frac{84 \text{ million tons}}{5} = 16.8 \text{ million tons}$

Thus rate of change is 16.8 million tons per year

5 0

3 years ago

Challenge: x(2)+9x+c :-D

kvv77 [185]

Answer:

×(2)+9×+c

2×+9×+c =

11x+c

3 0

3 years ago