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3 years ago

10

One weekend, a newsstand sold twice as many Sunday papers as Friday papers. The Sunday paper costs $1.50, the Friday paper costs

$0.75. How many Friday and Sunday papers were sold if the newsstand took in $116.25?

1 answer:

kicyunya [14]3 years ago

8 0

There was 51 Sunday papers and 53 Friday papers sold.

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Find the first 4 terms of the sequence.t(n) = -3n- 1b. t(15)=

aliya0001 [1]

Answer:

(a)-4, -7, -10 and -13.

(b) -46

Explanation:

Given the sequence:

$t\mleft(n\mright)=-3n-1$

(a) The first four terms are obtained by substituting n=1,2,3 and 4 respectively.

$\begin{gathered} t\mleft(1\mright)=-3(1)-1=-3-1=-4 \\ t(2)=-3(2)-1=-6-1=-7 \\ t(3)=-3(3)-1=-9-1=-10 \\ t(4)=-3(4)-1=-12-1=-13 \end{gathered}$

The first 4 terms of the sequence are -4, -7, -10 and -13.

(b)t(15)

$\begin{gathered} t(15)=-3(15)-1 \\ =-45-1 \\ =-46 \end{gathered}$

6 0

1 year ago

Please help me fast :)

liq [111]

Answer:

Step-by-step explanation:

Y = x² - 2x + 5

x = -2

y = -2² -2(-2) + 5

y = 4 + 4 + 5 = 13

x = -1

y = -1² -2(-1) + 5

y = 1 + 2 + 5 = 8

x = 1

y = 1² -2(1) + 5

y = 1 - 2 + 5 = 4

x = 3

y = 3² -2(3) + 5

y = 9 - 6 + 5 = 8

x = 6

y = 6² -2(6) + 5

y = 36 - 12 + 5 = 29

8 0

3 years ago

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

What is the area of this??? i put it at 14 points!

Basile [38]

Answer:

99in squared

Step-by-step explanation:

Bottom square is 3 by 3, middle is 9 by 6, and trapezoid is 9 by 4.

3*3+9*6+9*4 is 99

Hope this helps plz mark brainliest if correct :D

3 0

2 years ago

Guidelines for a wheelchair ramp suggest that the ratio of the rise to the run be no greater than 1:12.design a wheelchair ramp

OLEGan [10]

I have no idea tbh. I will try to find how to do it and help you!

4 0

3 years ago