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QveST [7]
2 years ago
5

If a×b÷2=a÷2 then b =​

Mathematics
1 answer:
lbvjy [14]2 years ago
8 0
If a×b÷2=a÷2 then b = 1
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Sonal went to school for 219 days in a full year. if her attendance is 75 percent find the number of days on which the school wa
Charra [1.4K]
Sonal went to school for 219 days in a full year. if her attendance is 75 percent find the number of days on which the school was open

7 0
3 years ago
PLEASE HELP ME IF YOU CAN! please tell me how you got the answer thank you!
Law Incorporation [45]

Answer:

D) 35.6

Step-by-step explanation:

Use pythagorean theorem

22^2 + 28^2 = 1268

Take square root of 1268, which equals to 35.60898763.

8 0
3 years ago
Help. MEEEEEEEEEEEEEEEEEE
a_sh-v [17]

Answer:

Step-by-step explanation:

2 Loaves of Bread is = 2.25*2 = 4.5 cups

She used the remaining flour to make 48 muffins.

48/24 = 2

3.25*2 = 6.5 cups

6.5 cups + 4.5 cups = 11 cups

6 0
2 years ago
Read 2 more answers
An instructor who taught two sections of Math 161A, the first with 20 students and the second with 30 students, gave a midterm e
uranmaximum [27]

Answer:

<em>The answers are for option (a) 0.2070  (b)0.3798  (c) 0.3938 </em>

Step-by-step explanation:

<em>Given:</em>

<em>Here Section 1 students = 20 </em>

<em> Section 2 students = 30 </em>

<em> Here there are 15 graded exam papers. </em>

<em> (a )Here Pr(10 are from second section) = ²⁰C₅ * ³⁰C₁₀/⁵⁰C₁₅= 0.2070 </em>

<em> (b) Here if x is the number of students copies of section 2 out of 15 exam papers. </em>

<em>  here the distribution is hyper-geometric one, where N = 50, K = 30 ; n = 15 </em>

<em>Then, </em>

<em> Pr( x ≥ 10 ; 15; 30 ; 50) = 0.3798 </em>

<em> (c) Here we have to find that at least 10 are from the same section that means if x ≥ 10 (at least 10 from section B) or x ≤ 5 (at least 10 from section 1) </em>

<em> so, </em>

<em> Pr(at least 10 of these are from the same section) = Pr(x ≤ 5 or x ≥ 10 ; 15 ; 30 ; 50) = Pr(x ≤ 5 ; 15 ; 30 ; 50) + Pr(x ≥ 10 ; 15 ; 30 ; 50) = 0.0140 + 0.3798 = 0.3938 </em>

<em> Note : Here the given distribution is Hyper-geometric distribution </em>

<em> where f(x) = kCₓ)(N-K)C(n-x)/ NCK in that way all these above values can be calculated.</em>

7 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
3 years ago
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