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lidiya [134]
4 years ago
6

I need help on this.​

Mathematics
2 answers:
ss7ja [257]4 years ago
7 0

Answer:

14

Step-by-step explanation:

DiKsa [7]4 years ago
7 0

Answer:

14

Step-by-step explanation:

hopefully this helps

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As school prefect, write a letter to your District or municipal or metropolitan executive requesting the maintenance of the buil
serious [3.7K]

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jsjsid sjdic8fnw jv 8djw efjiw eifwjiffjw djfivieo UK

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3 years ago
What is - 4/5 times -5/7
andreev551 [17]
What is - 4/5 times -5/7?
20/35 or in simplified form 4/7
6 0
3 years ago
Read 2 more answers
The probability that a male will be colorblind is .042. Find the probabilities that in a group of 53 men, the following are true
Dvinal [7]

Answer:

See below for answers and explanations

Step-by-step explanation:

<u>Part A</u>

\displaystyle P(X=x)=\binom{n}{x}p^xq^{n-x}\\\\P(X=5)=\binom{53}{5}(0.042)^5(1-0.042)^{53-5}\\\\P(X=5)=\frac{53!}{(53-5)!*5!}(0.042)^5(0.958)^{48}\\\\P(X=5)\approx0.0478

<u>Part B</u>

P(X\leq5)=P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)\\\\P(X\leq5)=\binom{53}{1}(0.042)^1(1-0.042)^{53-1}+\binom{53}{2}(0.042)^2(1-0.042)^{53-2}+\binom{53}{3}(0.042)^3(1-0.042)^{53-3}+\binom{53}{4}(0.042)^4(1-0.042)^{53-4}+\binom{53}{1}(0.042)^5(1-0.042)^{53-5}\\\\P(X\leq5)\approx0.9767

<u>Part C</u>

\displaystyle P(X\geq 1)=1-P(X=0)\\\\P(X\geq1)=1-(1-0.042)^{53}\\\\P(X\geq1)\approx1-0.1029\\\\P(X\geq1)\approx0.8971

6 0
3 years ago
A man invests his savings in two accounts ,one paying6% and the other paying 10% simple interest per year. He puts twice as much
AnnyKZ [126]

Answer: he invested $46062.5 at 6% and $23031.25 at 10%

Step-by-step explanation:

Let x represent the amount which he invested in the account paying 6% interest.

Let y represent the amount which he invested in the account paying 10% interest.

He puts twice as much in the lower-yielding account because it is less risky.. This means that

x = 2y

The formula for determining simple interest is expressed as

I = PRT/100

Considering the account paying 6% interest,

P = $x

T = 1 year

R = 6℅

I = (x × 6 × 1)/100 = 0.06x

Considering the account paying 10% interest,

P = $y

T = 1 year

R = 10℅

I = (y × 10 × 1)/100 = 0.1y

His annual interest is $7370dollars. it means that

0.06x + 0.2y = 7370 - - - - - - - - - -1

Substituting x = 2y into equation 1, it becomes

0.06 × 2y + 0.2y = 7370

0.12y + 0.2y = 7370

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y = 7370/0.32

y = $23031.25

x = 2 × 23031.25

x = 46062.5

6 0
3 years ago
Multiply radicals. Help with #52
timofeeve [1]

Answer:

\large\boxed{\sqrt{xy^3}\cdot\sqrt[3]{x^2y}=\sqrt[6]{x^7y^{11}}}

Step-by-step explanation:

\text{Use}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\\sqrt{xy^3}\cdot\sqrt[3]{x^2y}=(xy^3)^\frac{1}{2}(x^2y)^\frac{1}{3}\\\\\text{use}\ (ab)^n=a^nb^n\ \text{and}\ (a^n)^m=a^{nm}\\\\=x^\frac{1}{2}y^{(3)\left(\frac{1}{2}\right)}x^{(2)\left(\frac{1}{3}\right)}y^\frac{1}{3}\\\\\text{use}\ a^na^m=a^{n+m}\\\\=x^{\frac{1}{2}+\frac{2}{3}}y^{\frac{3}{2}+\frac{1}{3}}\\\\\text{the common denominator is 6}

\dfrac{1}{2}=\dfrac{1\cdot3}{2\cdot3}=\dfrac{3}{6}\\\\\dfrac{2}{3}=\dfrac{2\cdot2}{3\cdot2}=\dfrac{4}{6}\\\\\dfrac{3}{2}=\dfrac{3\cdot3}{2\cdot3}=\dfrac{9}{6}\\\\\dfrac{1}{3}=\dfrac{1\cdot2}{3\cdot2}=\dfrac{2}{6}\\\\x^{\frac{1}{2}+\frac{2}{3}}y^{\frac{3}{2}+\frac{1}{2}}=x^{\frac{3}{6}+\frac{4}{6}}y^{\frac{9}{6}+\frac{2}{6}}=x^{\frac{7}{6}}y^{\frac{11}{6}}=\sqrt[6]{x^7y^{11}}

5 0
4 years ago
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