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Elden [556K]
3 years ago
12

Please help hurry!

Mathematics
2 answers:
Natasha2012 [34]3 years ago
8 0

Answer:

  • ≈ 145.23 in²

Step-by-step explanation:

<u>The lateral surface area is:</u>

  • SA = 2πrh = 2*3.14*2.5*9.25 ≈ 145.23 in²
sdas [7]3 years ago
7 0
  • radius=2.5in=r
  • Height=h=9.25in=h

\\ \tt\longmapsto SA=2\pi rh[[/te][tex]\\ \tt\longmapsto SA=2\times 22}{7}\times (2.5)(9.25)

\\ \tt\longmapsto SA=145.23in^2

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For the given pair of events A and B, complete parts (a) and (b) below.
steposvetlana [31]

Answer:

(a) Option A. independent

(b) P(A\ and\ B) = 0.0312

Step-by-step explanation:

Two events A and B are considered dependent if the occurrence of one affects the probability of occurrence of another.

In this case, the events are independent because the birth of a girl or a boy does not affect the probability of obtaining a 1 by throwing a 16-sided die.

In the same way, throwing a die and obtaining the number 1 does not affect the probability that a baby is a girl or a boy.

When two events are independent then

P(A\ and\ B) = P(A)*P(B)

The probability that a baby is a girl is

P(A) = 0.5

The probability of obtaining a 1 when casting a 16-sided die is:

P(B) = \frac{1}{16}

Then:

P(A\ and\ B) = 0.5 * \frac{1}{16}

P(A\ and\ B) = 0.0312

5 0
3 years ago
Hello help me with this question thanks in advance​
Ede4ka [16]

\bold{\huge{\green{\underline{ Solutions }}}}

<h3><u>Answer </u><u>1</u><u>1</u><u> </u><u>:</u><u>-</u></h3>

<u>We </u><u>have</u><u>, </u>

\sf{HM = 5 cm }

  • <u>In </u><u>square </u><u>all </u><u>sides </u><u>of </u><u>squares </u><u>are </u><u>equal </u>

<u>The </u><u>perimeter </u><u>of </u><u>square </u>

\sf{ = 4 × side }

\sf{ = 4 × 5 }

\sf{ = 20 cm }

Thus, The perimeter of square is 20 cm

Hence, Option C is correct .

<h3><u>Answer </u><u>1</u><u>2</u><u> </u><u>:</u><u>-</u></h3>

<u>We </u><u>have</u><u>, </u>

\sf{MX  = 3.5 cm }

  • <u>In </u><u>square</u><u>,</u><u> </u><u>diagonals </u><u>are </u><u>equal </u><u>and </u><u>bisect </u><u>each </u><u>other </u><u>at </u><u>9</u><u>0</u><u>°</u>

<u>Here</u><u>, </u>

\sf{MX  = MT/2}

\sf{MT = 2 * 3.5 }

\sf{MT = 7 cm}

Thus, The MT is 7cm long

Hence, Option C is correct .

<h3><u>Answer </u><u>1</u><u>3</u><u> </u><u>:</u><u>-</u><u> </u></h3>

<u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>measure </u><u>of </u><u>Ang</u><u>l</u><u>e</u><u> </u><u>MAT</u>

  • <u>All </u><u>angles </u><u>of </u><u>square </u><u>are </u><u>9</u><u>0</u><u>°</u><u> </u><u>each </u>

<u>From </u><u>above </u>

\sf{\angle{MAT  = 90° }}

Thus, Angle MAT is 90°

Hence, Option B is correct .

<h3><u>Answer </u><u>1</u><u>4</u><u> </u><u>:</u><u>-</u></h3>

<u>We </u><u>know </u><u>that</u><u>, </u>

  • <u>All </u><u>the </u><u>angles </u><u>of </u><u>square </u><u>are </u><u>equal </u><u>and </u><u>9</u><u>0</u><u>°</u><u> </u><u>each </u>

<u>Therefore</u><u>, </u>

\sf{\angle{MHA  = }}{\sf{\angle{ MHT/2}}}

\sf{\angle{MHA = 90°/2}}

\sf{\angle {MHA = 45°}}

Thus, Angle MHA is 45°

Hence, Option A is correct

<h3><u>Answer </u><u>1</u><u>5</u><u> </u><u>:</u><u>-</u><u> </u></h3>

Refer the above attachment for solution

Hence, Option A is correct

<h3><u>Answer </u><u>1</u><u>6</u><u> </u><u>:</u><u>-</u><u> </u></h3>

Both a and b

  • <u>The </u><u>median </u><u>of </u><u>isosceles </u><u>trapezoid </u><u>is </u><u>parallel </u><u>to </u><u>the </u><u>base</u>
  • <u>The </u><u>diagonals </u><u>are </u><u>congruent </u>

Hence, Option C is correct

<h3><u>Answer </u><u>1</u><u>7</u><u> </u><u>:</u><u>-</u></h3>

In rhombus PALM,

  • <u>All </u><u>sides </u><u>and </u><u>opposite </u><u>angles </u><u>are </u><u>equal </u>

Let O be the midpoint of Rhombus PALM

<u>In </u><u>Δ</u><u>OLM</u><u>, </u><u>By </u><u>using </u><u>Angle </u><u>sum </u><u>property </u><u>:</u><u>-</u>

\sf{35° + 90° + }{\sf{\angle{ OLM = 180°}}}

\sf{\angle{OLM = 180° - 125°}}

\sf{\angle{ OLM = 55° }}

<u>Now</u><u>, </u>

\sf{\angle{OLM = }}{\sf{\angle{OLA}}}

  • <u>OL </u><u>is </u><u>the </u><u>bisector </u><u>of </u><u>diagonal </u><u>AM</u>

<u>Therefore</u><u>, </u>

\sf{\angle{ PLA = 55° }}

Thus, Angle PLA is 55° .

Hence, Option C is correct

8 0
2 years ago
What is the solution to
Anarel [89]

All my work is shown in the image.


4 0
3 years ago
Read 2 more answers
A financial advisor is analyzing a family's estate plan. The amount of money that the family has invested in different real esta
Pachacha [2.7K]

Answer:

The amount of money separating the lowest 80% of the amount invested from the highest 20% in a sampling distribution of 10 of the family's real estate holdings is $238,281.57.

Step-by-step explanation:

Let the random variable <em>X</em> represent the amount of money that the family has invested in different real estate properties.

The random variable <em>X</em> follows a Normal distribution with parameters <em>μ</em> = $225,000 and <em>σ</em> = $50,000.

It is provided that the family has invested in <em>n</em> = 10 different real estate properties.

Then the mean and standard deviation of amount of money that the family has invested in these 10 different real estate properties is:

\mu_{\bar x}=\mu=\$225,000\\\\\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{50000}{\sqrt{10}}=15811.39

Now the lowest 80% of the amount invested can be represented as follows:

P(\bar X

The value of <em>z</em> is 0.84.

*Use a <em>z</em>-table.

Compute the value of the mean amount invested as follows:

\bar x=\mu_{\bar x}+z\cdot \sigma_{\bar x}

   =225000+(0.84\times 15811.39)\\\\=225000+13281.5676\\\\=238281.5676\\\\\approx 238281.57

Thus, the amount of money separating the lowest 80% of the amount invested from the highest 20% in a sampling distribution of 10 of the family's real estate holdings is $238,281.57.

6 0
3 years ago
June bought three books. For the first book, she paid one half of her allowance plus $1 more. For the second book, she paid one
inysia [295]

Answer:

$34

Step-by-step explanation:

I'd work backwards.

For the third book, she paid all her remaining money. The problem says she paid "1/2 her leftover money + $3". This means that: (let m = money used to buy book 3)

m = 1/2m + 3

1/2m = 3

m = 6

For the second book: (let n = money before book 2)

n - m (money left after book 2) = 1/2n + 2

1/2n +2 is money used up for book, which is the same as n-m.

n = 1/2n + 2 + m

1/2n = 2 + m = 2 + 6

1/2n =8

n = 16

For the first book, she spent 1/2 her money + 1. If o = money before book 1 (or the whole allowance):

o - n = 1/2o + 1

o = 1/2o + n + 1 = 1/2o + 16 + 1

1/2o = 17

o = 34

Check!

Spent 17 (half 34) + 1 on book 1

16 left

Spent 8 (half 16) +2 on book 2

6 left

Spent 3 (half 6) + 3 on book 3

0 left

4 0
2 years ago
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