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2 years ago

Please answer this 22360 to 1 significant figure?

1 answer:

8 0

Zeros do not count as a sigfig so when you do something like this problem just include the numbers until you hit your significant figure limit and then substitute zeros where you can. 20000

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What are dot plots i’m confused

babymother [125]

A dot plot is a statistical chart consisting of data points plotted on a fairly simple scale, typically using filled in circles.

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2 years ago

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What measures of a cylinder can be determined when given the volume and the height? Describe how to find the measure.

Maru [420]

Answer:youywould be able to find the base.

Step-by-step explanation:

Formula for Volume of a cylinder is : V=Bh

8 0

2 years ago

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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

2 years ago

Use logarithms to solve each equation 2x9/8=111

Illusion [34]

<span>We have this equation:
</span>
$2x^{\frac{9}{8}} = 111$

and we need to find the value of x.

First of all, we multiply the whole equation for 1/2, so our goal is to isolate x, therefore:

$x^{\frac{9}{8}}=\frac{111}{2}$

Next step we must do is to apply <span>logarithms:
</span>
$logx^{\frac{9}{8}} = log(\frac{111}{2})$

Next, we have to apply identities and then to solve the equation:

$\frac{9}{8}logx = log(\frac{111}{2})$
$logx = \frac{8}{9}log(\frac{111}{2})$
$logx = 1.5504$
$10^{logx} = 10^{1.5504}$

Finally, we have the value of x which was our goal. This is the answer for the question above:

$x= 35.5140$

7 0

2 years ago

Which point is on the graph of the equation y = 2 x - 5?

Aneli [31]

Answer:

Step-by-step explanation:

the correct answer is number 4

if X = 2 ==> 2x -5 = 2(2) -5 = 4 - 5 = -1

4 0

2 years ago

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