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3 years ago

What is the answer I need ASAP

Mathematics

2 answers:

Ad libitum [116K]3 years ago

4 0

Answer:

Is that Egenuity? If it is I can go back and check my answer! d

Step-by-step explanation:

Diano4ka-milaya [45]3 years ago

3 0

I think it’s either d or b but I think it might be D

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Find the surface area of a square pyramid with a base length of 24 cm and a height of 16 cm. Explain how you solved the problem.

icang [17]

Answer:

the answer b

Step-by-step explanation:

[surface area]=[area of 4 triangles sides]+[area of square base]

[area of square base]=24*24--------> 576 cm²

[area of one triangles side]

l=slant height

l²=16²+12²-------> l²=400-----------> l=20 cm

[area of one triangles side]=24*20/2--------> 240 cm²

[area of 4 triangles sides]=4*240----------> 960 cm²

[surface area]=[960]+[576]--------> 1536 cm²

6 0

2 years ago

Solve Highschool Pre cal

Bezzdna [24]

Option C:

$\ln 64 y^{2}$

Solution:

Given expression: 2 ln 8 + 2 ln y

Applying log rule $a \log b=\log b^{a}$ in the above expression.

⇒ $2 \ln 8+2 \ln y=\ln 8^{2}+\ln y^{2}$

Applying log rule $\log a+\log b=\log (a b)$ in the above expression.

⇒ $\ln 8^{2}+\ln y^{2}=\ln \left(8^{2} y^{2}\right)$

We know that $8^{2}=64$

⇒ $\ln \left(8^{2} y^{2}\right)=\ln 64 y^{2}$

Hence, the expression in single natural logarithm is $\ln 64 y^{2}$ .

6 0

3 years ago

Simplify: (32x2/5y5)1/5

masya89 [10]

The answer is 64/5 xy

4 0

3 years ago

jim has a dog. jim buys 6 bags of dog biscuits. there are 27 biscuits in each bag of dog biscuits. jims dog eats 5 dog biscuits

Nonamiya [84]

Answer:

i will give it to you

Step-by-step explanation:

that's it for you dear

7 0

2 years ago

You have a large jar that initially contains 30 red marbles and 20 blue marbles. We also have a large supply of extra marbles of

Dima020 [189]

Answer:

There is a 57.68% probability that this last marble is red.

There is a 20.78% probability that we actually drew the same marble all four times.

Step-by-step explanation:

Initially, there are 50 marbles, of which:

30 are red

20 are blue

Any time a red marble is drawn:

The marble is placed back, and another three red marbles are added

Any time a blue marble is drawn

The marble is placed back, and another five blue marbles are added.

The first three marbles can have the following combinations:

R - R - R

R - R - B

R - B - R

R - B - B

B - R - R

B - R - B

B - B - R

B - B - B

Now, for each case, we have to find the probability that the last marble is red. So

$P = P_{1} + P_{2} + P_{3} + P_{4} + P_{5} + P_{6} + P_{7} + P_{8}$

$P_{1}$ is the probability that we go R - R - R - R

There are 50 marbles, of which 30 are red. So, the probability of the first marble sorted being red is $\frac{30}{50} = \frac{3}{5}$ .

Now the red marble is returned to the bag, and another 3 red marbles are added.

Now there are 53 marbles, of which 33 are red. So, when the first marble sorted is red, the probability that the second is also red is $\frac{33}{53}$

Again, the red marble is returned to the bag, and another 3 red marbles are added

Now there are 56 marbles, of which 36 are red. So, in this sequence, the probability of the third marble sorted being red is $\frac{36}{56}$

Again, the red marble sorted is returned, and another 3 are added.

Now there are 59 marbles, of which 39 are red. So, in this sequence, the probability of the fourth marble sorted being red is $\frac{39}{59}$ . So

$P_{1} = \frac{3}{5}*\frac{33}{53}*\frac{36}{56}*\frac{39}{59} = \frac{138996}{875560} = 0.1588$

$P_{2}$ is the probability that we go R - R - B - R

$P_{2} = \frac{3}{5}*\frac{33}{53}*\frac{20}{56}*\frac{36}{61} = \frac{71280}{905240} = 0.0788$

$P_{3}$ is the probability that we go R - B - R - R

$P_{3} = \frac{3}{5}*\frac{20}{53}*\frac{33}{58}*\frac{36}{61} = \frac{71280}{937570} = 0.076$

$P_{4}$ is the probability that we go R - B - B - R

$P_{4} = \frac{3}{5}*\frac{20}{53}*\frac{25}{58}*\frac{33}{63} = \frac{49500}{968310} = 0.0511$

$P_{5}$ is the probability that we go B - R - R - R

$P_{5} = \frac{2}{5}*\frac{30}{55}*\frac{33}{58}*\frac{36}{61} = \frac{71280}{972950} = 0.0733$

$P_{6}$ is the probability that we go B - R - B - R

$P_{6} = \frac{2}{5}*\frac{30}{55}*\frac{25}{58}*\frac{33}{63} = \frac{49500}{1004850} = 0.0493$

$P_{7}$ is the probability that we go B - B - R - R

$P_{7} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{33}{63} = \frac{825}{17325} = 0.0476$

$P_{8}$ is the probability that we go B - B - B - R

$P_{8} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{30}{65} = \frac{750}{17875} = 0.0419$

So, the probability that this last marble is red is:

$P = P_{1} + P_{2} + P_{3} + P_{4} + P_{5} + P_{6} + P_{7} + P_{8} = 0.1588 + 0.0788 + 0.076 + 0.0511 + 0.0733 + 0.0493 + 0.0476 + 0.0419 = 0.5768$

There is a 57.68% probability that this last marble is red.

What's the probability that we actually drew the same marble all four times?

$P = P_{1} + P_{2}$

$P_{1}$ is the probability that we go R-R-R-R. It is the same $P_{1}$ from the previous item(the last marble being red). So $P_{1} = 0.1588$

$P_{2}$ is the probability that we go B-B-B-B. It is almost the same as $P_{8}$ in the previous exercise. The lone difference is that for the last marble we want it to be blue. There are 65 marbles, 35 of which are blue.

$P_{2} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{35}{65} = \frac{875}{17875} = 0.0490$

$P = P_{1} + P_{2} = 0.1588 + 0.0490 = 0.2078$

There is a 20.78% probability that we actually drew the same marble all four times

3 0

3 years ago