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4 years ago

A card is drawn from a well-shuffled deck of 52. Find p(drawing a face or a 4)

Mathematics

1 answer:

Musya8 [376]4 years ago

4 0

Answer:

the answer is none of those it is 4/13

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Kristen has been earning $150 per week by babysitting. She has been working hard to save 20% of each week's earnings. If she con

anyanavicka [17]

Answer:

it will take her 6 weeks

Step-by-step explanation:

150*.2=30

180/30=6

5 0

4 years ago

Facts about pasical's education

Paha777 [63]

Blaise Pascal (June 19, 1623 - August 19, 1662) was a great contributor to math, science, and philosophy, especially Christian philosophy. Interesting Blaise Pascal Facts: Pascal's early education in France was conducted at home by his father due to the prodigious talent and understanding he showed as a child. Did u mean pascal instead of pasical becaus ethere is no such thing. Hope I helped :)

4 0

3 years ago

Which of the following expressions is equivalent to (3x+6)(2x-1)

Elden [556K]

Answer:

6x^2+9x-6

Step-by-step explanation:

(3x+6)(2x-1)=6x^2+12x-3x-6=6x^2+9x-6

4 0

3 years ago

In a class of 40 students, everyone has either a pierced nose or a pierced ear. The professor asks everyone with a pierced nose

viktelen [127]

Answer:

3 students

Step-by-step explanation:

If everyone in the class has either a pierced nose or ear, we just simply have to add up the total number of hands raised and minus the number of students in the class.

35+8=43

43-40=3

3 students have both a pierced nose and pierced ear.

6 0

3 years ago

Given: △ACM, m∠C=90°, CP ⊥ AM

larisa86 [58]

Answer: The answer is $3\dfrac{4}{7}.$

Step-by-step explanation: Given in the question that ΔAM is a right-angled triangle, where ∠C = 90°, CP ⊥ AM, AC : CM = 3 : 4 and MP - AP = 1. We are to find AM.

Let, AC = 3x and CM = 4x.

In the right-angled triangle ACM, we have

$AM^2=AC^2+CM^2=(3x)^2+(4x)^2=9x^2+16x^2=25x^2\\\\\Rightarrow AM=5x.$

Now,

$AM=AP+PM=AP+(AP+1)\\\\\Rightarrow 2AP=AM-1\\\\\Rightarrow 2AP=5x-1.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(A)$

Now, since CP ⊥ AM, so ΔACP and ΔMCP are both right-angled triangles.

So,

$CP^2=AC^2-AP^2=CM^2-MP^2\\\\\Rightarrow (3x)^2-AP^2=(4x)^2-(AP+1)^2\\\\\Rightarrow 9x^2-AP^2=16x^2-AP^2-2AP-1\\\\\Rightarrow 2AP=7x^2-1.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(B)$

Comparing equations (A) and (B), we have

$5x-1=7x^2-1\\\\\Rightarrow 5x=7x^2\\\\\Rightarrow x=\dfrac{5}{7},~\textup{since }x\neq 0.$

Thus,

$AM=5\times\dfrac{5}{7}=\dfrac{25}{7}=3\dfrac{4}{7}.$

8 0

4 years ago