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3 years ago

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Ted likes to run long distances. He can run 20 \text{ km}20 km20, start text, space, k, m, end text in 959595 minutes. He wants

to know how many kilometers (k)(k)left parenthesis, k, right parenthesis he will go if he runs at the same pace for 285285285 minutes.

1 answer:

Ghella [55]3 years ago

5 0

Answer:

im looking for the anwser

Step-by-step explanation:

You might be interested in

HElPPPPPPPPP PPZPZLXPLZPLZPZZZZZZZZ

monitta

Answer:

5. 43 / 6

6. 8.027778

2. -0.167

4. 1.5

Step-by-step explanation:

5. Conversion a mixed number to a improper fraction: 9 2/3 = 9 · 3 + 2 /3 = 29/3

Conversion a mixed number to a improper fraction: 2 1/2 = 2 · 2 + 1 /2 = 5/2

Subtract: 29/3 - 5/2 = 29 · 2 / 3 · 2 - 5 · 3 / 2 · 3 = 58 / 6 - 15 / 6 = 58 - 15 / 6 = 43 / 6

6. 2 4/9 + 5 7/12 = 289 / 36 = 81 / 36 = 8.027778

2. 3/4 - 11/12 = -1 / 6 =-0.1666667

4. 1 9/10 - 2/5 = 3 / 2 = 11 / 2 = 1.5

5 0

3 years ago

When seven times a number is deceased by 3, the result is 25. What is the number?

zepelin [54]

7n-3= 25

7n-3(+3)= 25(+3)

(7n= 28)/7

n=4

6 0

3 years ago

Read 2 more answers

For two events A and B show that P (A∩B) ≥ P (A)+P (B)−1.

nordsb [41]

Answer:

<h3>For two events A and B show that P (A∩B) ≥ P (A)+P (B)−1.</h3>

By De morgan's law

$(A\cap B)^{c}=A^{c}\cup B^{c}\\\\P((A\cap B)^{c})=P(A^{c}\cup B^{c})\leq P(A^{c})+P(B^{c}) \\\\1-P(A\cap B)\leq P(A^{c})+P(B^{c}) \\\\1-P(A\cap B)\leq 1-P(A)+1-P(B)\\\\-P(A\cap B)\leq 1-P(A)-P(B)\\\\P(A\cap B)\geq P(A)+P(B)-1$

which is Bonferroni’s inequality

<h3>Result 1: P (Ac) = 1 − P(A)</h3>

Proof

If S is universal set then

$A\cup A^{c}=S\\\\P(A\cup A^{c})=P(S)\\\\P(A)+P(A^{c})=1\\\\P(A^{c})=1-P(A)$

<h3>Result 2 : For any two events A and B, P (A∪B) = P (A)+P (B)−P (A∩B) and P(A) ≥ P(B)</h3>

Proof:

If S is a universal set then:

$A\cup(B\cap A^{c})=(A\cup B) \cap (A\cup A^{c})\\=(A\cup B) \cap S\\A\cup(B\cap A^{c})=(A\cup B)$

Which show A∪B can be expressed as union of two disjoint sets.

If A and (B∩Ac) are two disjoint sets then

$P(A\cup B) =P(A) + P(B\cap A^{c})---(1)\\$

B can be expressed as:

$B=B\cap(A\cup A^{c})\\$

If B is intersection of two disjoint sets then

$P(B)=P(B\cap(A)+P(B\cup A^{c})\\P(B\cup A^{c}=P(B)-P(B\cap A)$

Then (1) becomes

$P(A\cup B) =P(A) +P(B)-P(A\cap B)\\$

<h3>Result 3: For any two events A and B, P(A) = P(A ∩ B) + P (A ∩ Bc)</h3>

Proof:

If A and B are two disjoint sets then

$A=A\cap(B\cup B^{c})\\A=(A\cap B) \cup (A\cap B^{c})\\P(A)=P(A\cap B) + P(A\cap B^{c})\\$

<h3>Result 4: If B ⊂ A, then A∩B = B. Therefore P (A)−P (B) = P (A ∩ Bc) </h3>

Proof:

If B is subset of A then all elements of B lie in A so A ∩ B =B

$A =(A \cap B)\cup (A\cap B^{c}) = B \cup ( A\cap B^{c})$

where A and A ∩ Bc are disjoint.

$P(A)=P(B\cup ( A\cap B^{c}))\\\\P(A)=P(B)+P( A\cap B^{c})$

From axiom P(E)≥0

$P( A\cap B^{c})\geq 0\\\\P(A)-P(B)=P( A\cap B^{c})\\P(A)=P(B)+P(A\cap B^{c})\geq P(B)$

Therefore,

P(A)≥P(B)

8 0

3 years ago

A 3 inch by 5 inch photograph is enlarged so that the longer side is 15 inches.

valentina_108 [34]

Answer:

the length of shorter side is 9 inches.

longer side is enlarged 3times

so shorter side will also be enlarged 3 times.

5 0

3 years ago

Question is attached below, Good Luck! I will Mark Brainliest for best answer!

DerKrebs [107]

Answer:

1960 cm^2

Step-by-step explanation:

Surface area:

Area of cross section x 2 = 1/2 x 20 x 21 x 2 = 420 cm^2

Area of slope = 29 x 22 = 638cm^2

Area of base = 22 x 20 = 440cm^2

Area of back = 21 x 22 = 462cm^2

Total Surface area = 462+440+638+420 = 1960cm^2

Hope this helps

6 0

3 years ago