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morpeh [17]
3 years ago
11

Ok I need this question ASAP

Mathematics
1 answer:
zysi [14]3 years ago
8 0

Answer:

Volume = 2960 ft^3

Step-by-step explanation:

Thanks!

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14 bugs are crawling on the steps. Some of the bugs are black, and 4 are
docker41 [41]

Answer:

b  = 14-4

Step-by-step explanation:

Let the number of black bugs be b

Let the number of green bugs be g

If 14 bugs are crawling on the step, then;

b + g = 14 ....1

If there are 4 green bugs, then g = 4

Substitute g = 4 into the equation

b + g = 14

b + 4 = 14

b = 14 - 4

Hence the sentence that can be required to find the number of black bugs is b  = 14-4

8 0
3 years ago
Simplify:2 1/3(4 1/4÷2)
kondor19780726 [428]

Answer:

4\frac{23}{24}

Step-by-step explanation:

We want to simplify:

2\frac{1}{3}(4\frac{1}{4}\div 2)

We convert to improper fraction to get:

\frac{7}{3}(\frac{17}{4}\div 2)

We now deal with the parenthesis by multiplying with the reciprocal of 2.

\frac{7}{3}(\frac{17}{4}\times \frac{1}{2} )

This gives:

\frac{7}{3}(\frac{17}{8})

We multiply to get:

\frac{7}{3}(\frac{17}{8})

This simplfies to:

\frac{119}{24}

We write as mixed number

4\frac{23}{24}

4 0
3 years ago
8th grade math
UkoKoshka [18]

Answer:

10 Milk chocolates

8 dark chocolates

6 white chocolates

24 TOTAL

\frac{10}{24} Probability that Hanissa choose a milk chocolate

Now, since she has already eaten one chocolate, then what's left is 23 chocolates.

So, since the second one should be a white chocolate and there are 6 of them, then

\frac{6}{23} - probability of choosing a white chocolate.

Then, multiply,

( \frac{10}{24} )( \frac{6}{23} )= \frac{60}{552} or \frac{5}{46}

Answer: 5/46 or 10.87%

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
‼️HELEP MEEE⛔️<br><br> ⁉️SOS
Zolol [24]

\\ \sf\longmapsto x+40=90(Complementary)

\\ \sf\longmapsto x=90-40

\\ \sf\longmapsto x=50

Now

\\ \sf\longmapsto y+40=180

\\ \sf\longmapsto y=180-40

\\ \sf\longmapsto y=140

Now

\\ \sf\longmapsto x+y=140+50=190

7 0
3 years ago
True or False? The sum of the differences must be zero for any distribution consisting of n observations.
MakcuM [25]
Its TRUE!!! :) Hope this helps.
3 0
3 years ago
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