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3 years ago

23. The density of mercury is 13g/cm cubed. If there are 4 cm cubed of mercury, what is the mass?

Mathematics

2 answers:

ZanzabumX [31]3 years ago

6 0

Answer:

Step-by-step explanation:

zhenek [66]3 years ago

5 0

I believe it would be something related to the corresponding numbers.

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A total of 31,000 is to be invested, some in bonds and some in certificates of deposit (CDs). If the amount invested in bonds is

kupik [55]

Answer:

Bonds: $42,000

Certificates of deposit: $41,000

Step-by-step explanation:

Total invested = Amount in bonds + Amount in CDs

Amount in bonds = Amount in CDs + 1000

Let the amount in bonds = B and the amount in CDs = C

1. 83,000 = B + C

2. B = C+1000

Since the above expression (#2) defines B, you can substitute it for the B in the first equation (#1).

83,000 = C + 1000 + C

Now, you can solve for C.

83,000 = 2C + 1000

82,000 = 2C

41,000 = C

You know that the amount invested in bonds is $1000 greater than the amount invested in CDs, so add $1000 to C and you find B, $42,000.

4 0

3 years ago

HELP PLEASE THANK YOU! :) 50 POINTS !!!!!!!1

a_sh-v [17]

Answer: 0.88

Step-by-step explanation:

Let C is the event of drinking coffee, T is the event of drinking tea and M is the event of drinking milk.

Thus, when we make the Venn diagram of the given situation according to the given information,

Total number of people = 50

Number of people who like coffee, tea and milk = 19

Number of people who like coffee, tea but not milk = 16

Number of people who like coffee, milk but not tea = 2

Number of people who like tea, milk but not coffee = 5

Thus, the number of people who like tea only = Total people - (people who like coffee, tea but not milk + people who like coffee, tea and milk + the one who only like tea and milk but not coffee)

= 50 - ( 16 + 19 + 5) = 50 - 46 = 4

Thus, Total number of the person who like milk = 16 + 19 + 5 + 4 = 44

⇒ Probability that this person likes tea = $\frac{\text{ Total person who like to drink milk}}{\text{ Total number of people}}$ = $\frac{44}{50} = 0.88$

3 0

3 years ago

Determine the sum of the arithmetic series: 5+18 +31 +44 + ... 161.

Finger [1]

Answer:

1079

Step-by-step explanation:

Hello,

18-5 = 13

31-18=13

44-31=13

161=5+13*12

So we need to compute

$\displaystyle \sum_{k=0}^{k=12} \ {(5+13k)}\\\\=\sum_{k=0}^{k=12} \ {(5)} + 13\sum_{k=1}^{k=12} \ {(k)}\\\\=13*5+13*\dfrac{12*13}{2}\\\\=65+13*13*6\\\\=65+1014\\\\=1079$

Thanks

3 0

3 years ago

Please answer :puppy eyes emoji:

Anettt [7]

Well ven diagrams help you visually compare and contrast two figures. They helps you note the similarities and the differences between two items. Hope this helps!

8 0

3 years ago

In a group of a hundred and fifty students attending a youth workshop in mombasa, 125 of them are fluent in kiswahili, 135 in en

jek_recluse [69]

Answer:

The probability that a student chosen at random is fluent in English or Swahili.

P(S∪E) = 1.1

Step-by-step explanation:

Step(i):-

Given total number of students n(T) = 150

Given 125 of them are fluent in Swahili

Let 'S' be the event of fluent in Swahili language

n(S) = 125

The probability that the fluent in Swahili language

$P(S) = \frac{n(S)}{n(T)} = \frac{125}{150} = 0.8333$

Let 'E' be the event of fluent in English language

n(E) = 135

The probability that the fluent in English language

$P(E) = \frac{n(E)}{n(T)} = \frac{135}{150} = 0.9$

n(E∩S) = 95

The probability that the fluent in English and Swahili

$P(SnE) = \frac{n(SnE)}{n(T)} = \frac{95}{150} = 0.633$

Step(ii):-

The probability that a student chosen at random is fluent in English or Swahili.

P(S∪E) = P(S) + P(E) - P(S∩E)

= 0.833+0.9-0.633

= 1.1

Final answer:-

The probability that a student chosen at random is fluent in English or Swahili.

P(S∪E) = 1.1

8 0

3 years ago