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3 years ago

Hey can you please help me posted picture of question

Mathematics

1 answer:

Nataliya [291]3 years ago

3 0

Total number of blue candies = 20 + 30 = 50
Number of blue candies without nut = 20

Probability of picking a blue candy without a nut = (Number of blue candies with nut) /(Total number of blue candies)

So, Probability of picking a blue candy without a nut = $\frac{20}{50} =0.4=40$ %

Thus, the correct answer is option A

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If you could answer both, that would be amazing:)

sveticcg [70]

The first question is “32.4”
I’ll come back if I find the other

7 0

3 years ago

↓↓ASAP PLAS HELP WILL GIVE BRAINLYEST WILL GIVE BRAINLYEST↓↓

Stels [109]

we know that

diameter of the base of a cylindrical can= 4 in----------> r=2 in

h=6.5 in

[surface area]=2*[pi*r²]+[2*pi*r*h]

then

[surface area]=2*[pi*2²]+[2*pi*2*6.5]=[ 25.13 ]+[ 81.68 ]=106.81 in²

the answer is 106.8 in²

3 0

3 years ago

Consider all 5 letter "words" made from the full English alphabet. (a) How many of these words are there total? (b) How many of

VARVARA [1.3K]

Answer:

a) There are 11,881,336 of these words in total.

b) There are 7,893,600 of these words with no repeated letters.

c) 896,376 of these words start with an a or end with a z or both

Step-by-step explanation:

Our words have the following format:

L1 - L2 - L3 - L4 - L5

In which L1 is the first letter, L2 the second letter, etc...

There are 26 letters in the English alphabet.

(a) How many of these words are there total?

Each of L1, L2, L3, L4 and L5 have 26 possible options.

So there are $26^{5} = 11,881,336$ of these words total

(b) How many of these words contain no repeated letters?

The first letter can be any of them, so L1 = 26.

At the second letter, the first one cannot be repeated, so L2 = L1 - 1 = 25.

At the third letter, nor the first nor the second one can be repeated, so L3 = L1 - 2 = 24

This logic applies until L5

So we have

26-25-24-23-22

In total there are

$26*25*24*23*22 = 7,893,600$

of these words with no repeated letters.

(c) How many of these words start with an a or end with a z or both (repeated letters are allowed)?

$T = T_{1} + T_{2} + T_{3}$

$T_{1}$ is the number of words that start with an a and do not end with z. So we have

1 - 26 - 26 - 26 - 25

The first letter can only be a, and the last one cannot be z. So:

$T_{1} = 26^{3}*25 = 439,400$

$T_{2}$ is the number of words that start with any letter other than a and end with z. So we have

25 - 26 - 26 - 25 - 1

The first letter can be any of them, other than a, and the last can only be z. So:

$T_{2} = 26^{3}*25 = 439,400$

$T_{3}$ is the number of words that both start with a and end with z. So:

1 - 26 - 26 - 26 - 1

The first letter can only be a, and the last can only be z. The other three letters could be anything. So:

$T_{3} = 26^{3} = 17,576$

$T = T_{1} + T_{2} + T_{3} = 2*439,400 + 17,576 = 896,376$

896,376 of these words start with an a or end with a z or both

4 0

3 years ago

Find the sum of the arithmetic sequence. 3,5,7,9,....,21

inysia [295]

The sum of any arithmetic sequence is the average of the first and last terms times the number of terms.

Any term in an arithmetic sequence is:

a(n)=a+d(n-1), where a=initial term, d=common difference, n=term number

So the first term is a, and the last term is a+d(n-1) so the sum can be expressed as:

s(n)(a+a+d(n-1))(n/2)

s(n)=(2a+dn-d)(n/2)

s(n)=(2an+dn^2-dn)/2

However we need to know how many terms are in the sequence.

a(n)=a+d(n-1), and we know a=3 and d=2 and a(n)=21 so

21=3+2(n-1)

18=2(n-1)

9=n-1

10=n so there are 10 terms in the sequence.

s(n)=(2an+dn^2-dn)/2, becomes, a=3, d=2, n=10

s(10)=(2*3*10+2*10^2-2*10)/2

s(10)=(60+200-20)/2

s(10)=240/2

s(10)=120

7 0

3 years ago

One student says -5 is bigger than -4 and uses money as the analogy: “if i owe $5, i have a bigger debt than if i owe $4.” what

makkiz [27]

Answer:

-5 is smaller than -4 bc -4 is closer to zero meaning it has more value than 5

Step-by-step explanation:

8 0

3 years ago