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Usimov [2.4K]
3 years ago
11

Q and R are independent events. P(Q) = 0.2; P(Q and R) = 0.09. Find P(R)

Mathematics
1 answer:
nasty-shy [4]3 years ago
3 0
Is it P (q+r ) or P( q, r)?
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A rectangular park is 100 yards long and 65 yards wide. Give the length and width of another rectangular park that has the same
AlekseyPX
2* 100 + 2 * 65 = 330 perimeter
100 * 65 = 6,500 area

2*90 + 2*75 = 330 perimeter
90 * 75 = 6,750 area


8 0
3 years ago
Solve step by step solution then only i can do it plxx ​
Anuta_ua [19.1K]

Answer:

<h3><u>Let's</u><u> </u><u>understand the concept</u><u>:</u><u>-</u></h3>

Here angle B is 90°

So \triangle ABC and \triangle ABD Are right angled triangle

So we use Pythagoras thereon for solution

<h3><u>Required Answer</u><u>:</u><u>-</u></h3>
  • First in triangle ABC

perpendicular=p=8cm

Hypontenuse =h =10cm

  • We need to find base=b

According to Pythagoras thereon

{\boxed{\sf b^2=h^2-p^2}}

  • Substitutethe values

\longrightarrow\sf b^2=10^2-p^2

\longrightarrow\sf b={\sqrt {10^2-8^2}}

\longrightarrow\sf b={\sqrt{100-64}}

\longrightarrow\bf b={\sqrt {36}}

\longrightarrow\sf b=6

\therefore\overline{BC}=6cm

  • BD=BC+CD

\longrightarrowBD=9+6

\longrightarrowBD=15cm

  • Now in \triangle ABD

Perpendicular=p=8cm

Base =b=15cm

  • We need to find Hypontenuse =AD(x)

According to Pythagoras thereon

{\boxed {\sf h^2=p^2+b^2}}

  • Substitute the values

\longrightarrow\sf h^2=8^2+15^2

\longrightarrow\sf h={\sqrt {8^2+15^2}}

\longrightarrow\sf h={\sqrt {64+225}}

\longrightarrow\sf h={\sqrt {289}}

\longrightarrow\sf h=17cm

\therefore{\underline{\boxed{\bf x=17cm}}}

3 0
2 years ago
I need help for solving 4.8(x+4)=2.16
Bas_tet [7]
I hope this helps you



4,8 (x+4)=2,16



48 (x+4)=21,6


x+4=21,6/48


x+4= 0,45


x=0,45-4


x= -3,55


5 0
3 years ago
Read 2 more answers
It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per nig
Kitty [74]

Answer: No , at 0.05 level of significance , we have sufficient evidence to reject the claim that LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average.

Step-by-step explanation:

Let \mu denotes the average hours of sleep per night.

As per given , we have

H_0:\mu=7\\H_a:\mu

, since H_a is left-tailed and population standard deviation is unknown, so the test is a left-tailed t -test.

Also , it is given that ,

Sample size : n= 22

Sample mean : \overline{x}=7.24

Sample  standard deviation : s= 1.93

Test statistic : t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}

i.e.  t=\dfrac{7.24-7}{\dfrac{1.93}{\sqrt{22}}}\approx0.58

For significance level \alpha=0.05 and degree of freedom 21 (df=n-1),

Critical t-value for left-tailed test= t_{\alpha, df}=t_{0.05,21}=- 1.7207

Decision : Since the test statistic value (0.58) > critical value  1.7207, it means we are failed to reject the null hypothesis .

[Note : When |t_{cal}|>|t_{cri}|, then we accept the null hypothesis.]

Conclusion: We have sufficient evidence to reject the claim that LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average.

6 0
2 years ago
Mrs.mayfields class sold a total of 7293 lollipops for 2.00 each they bought the lollipops for 1.00 each. how much profit did th
Akimi4 [234]
$7293 I think....

7293 x 2 = 14586 / 2 = 7293
6 0
3 years ago
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