# A newspaper’s cover page is 3/8 text, and photographs fill the rest. If 2/5 of the text is an article aboutendangered species, what fraction of the cover page is the article about endangered species

Step-by-step explanation:

In this 2/5 of 3/8 is about endangered species

Therefore to obtain the answer we need to multiply the two fractions

2/5 × 3/8

Fraction of the cover page about endangered species is therefore = 2/5 × 3/8

= 0.15

## Related Questions

1. 2x2 (3+8)

Step-by-step explanation:

Because (3+8) is in paranthesis, you should do this first: 2x2(11)

Now, because there is no sign between the second 2 and the parantheses, this simply means they are multiplied so now you have: 2x2x11

Now, because they all have the same sign, you can just multiply across:

2x2x11 = 44

What to do for the me part ​

Me: $100,00 Maximum total debt:$20,000
Maximum monthly payments: $833.30 Find the length of the indicated segments of each trapezoid. ### Answers Answer: BC = 16 EF = 23 AD = 30 Step-by-step explanation: Evaluate the given integral by making an appropriate change of variables, where R is the region in the first quadrant bounded by the ellipse 64x2 + 81y2 = 1.$ L=\iint_{R} {\color{red}9} \sin ({\color{red}384} x^{2} + {\color{red}486} y^{2})\,dA \$.

Notice that Given that is an ellipse, consider a conversion to polar coordinates:

The Jacobian for this transformation is

with determinant

Then the integral in polar coordinates is

where you can evaluate the remaining integral by substituting and .

To evaluate the integral, we make a change of variables using the transformation x=u/8 and y=v/9 to transform the region into a unit circle. Then we convert the integral to polar coordinates and evaluate it.

### Explanation:

To evaluate the given integral, we can make the appropriate change of variables by using the transformation x = u/8 and y = v/9. This will transform the region R into a unit circle. The determinant of the Jacobian of the transformation is 1/72, which we will use to change the differential area element from dA to du dv. Substituting the new variables and limits of integration, the integral becomes:

L = \iint_{R} 9 \sin (612 u^{2} + 768 v^{2}) \cdot (1/72) \,du \,dv

Next, we can convert the integral from Cartesian coordinates(u, v) to polar coordinates (r, \theta). The integral can be rewritten as:

L = \int_{0}^{2\pi} \int_{0}^{1} 9 \sin (612 r^{2} \cos^{2}(\theta) + 768 r^{2} \sin^{2}(\theta)) \cdot (1/72) \cdot r \,dr \,d\theta

We can then evaluate this integral to find the value of L.

brainly.com/question/32205191

#SPJ11

Every day a kindergarten class chooses randomly one of the 50 state flags to hang on the wall, without regard to previous choices, We are interested in the flags that are chosen on Monday, Tuesday and Wednesday of next week.a) Describe a sample space \Omega and a probability measure P to model this experiment.

b) What is the probability that the class hangs Wisconsin's flag on Monday, Michigan's flag on Tuesday, and California's flag on Wednesday.?

c) What is the probability that Wisconsin's flag will be hung at least two of the three days?

a.)  P(x = X) =

b.)

c.) 0.00118

Step-by-step explanation:

The sample space Ω = flags of all 50 states

a.) Any one of the flags is randomly chosen. Therefore we can write the

probability measure as P(x = X) = , for all the elements of the sample

space, that is for all x ∈ Ω.

b.) the probability that the class hangs Wisconsin's flag on Monday,

Michigan's flag on Tuesday, and California's flag on Wednesday

=

c.) the probability that Wisconsin's flag will be hung at least two of the three days

= Probability that Wisconsin's flag will be hung on two days + Probability that Wisconsin's flag will be hung on three days

= P(x = 2) + P(x = 3)

=

=

=

= 0.00118

The sample space for this experiment is all the possible combinations of flags from the 50 U.S. states for the three days. The probability of hanging Wisconsin's flag on Monday, Michigan's on Tuesday, and California's on Wednesday is 1/125,000. The probability of hanging Wisconsin's flag at least two of the three days is 294/125,000.

### Explanation:

a) The sample space Ω for this experiment comprises of all possible combinations of flags from the 50 U.S. states for the three days. Hence, the total number of outcomes in the sample space Ω would be 50*50*50 = 125,000. Every outcome in this space is equally likely, so the probability measure P would assign a probability of 1/125,000 to each outcome.

b) As each day's choice is independent of the others and each state's flag is equally likely to be chosen, the probability that Wisconsin's flag is hung on Monday, Michigan's flag is hung on Tuesday, and California's flag is hung on Wednesday would be (1/50) * (1/50) * (1/50) = 1/125,000.

c) To find the probability that Wisconsin's flag will be hung at least two of the three days, we have to add the probabilities for the three situations where Wisconsin's flag is hung exactly twice plus the situation where Wisconsin's flag is hung all three days. The final probability would be [(3 * (1/50)² * (49/50)) + (1/50)³] = 294/125,000.

brainly.com/question/22962752

#SPJ3

B. 1/2 of 4/5 = ____ fifth(s)

2 because of the way the question is written

2/5 if you get 2 wrong.

Step-by-step explanation:

1/2 * 4/5  of means multiply

numerator:  1*4 = 4

denominator = 2 * 5 = 10

Answer: 4 / 10 but this will reduce. Divide Top and Bottom by 2

4/2 = 2

10/2 = 5