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lina2011 [118]
4 years ago
7

How do I solve this

Mathematics
1 answer:
Ostrovityanka [42]4 years ago
6 0

Answer: the fourth one

Step-by-step explanation:

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Helppppppppppppppppppppppppp
Wewaii [24]

Answer:

Radius: 8

Circumference: 50.27

πr^2 = Area of circle

201.06 = πr^2

divide by pi

201.06 / π = 63.9993857161 = 64

64 = r^2

Square root both sides to get radius.

\sqrt{64} = 8

r = 8

Circumference = π x diameter

Diameter = 2 x radius

Circumference = π x 16 = 50.2654824574 = 50.27

8 0
3 years ago
Help please quick
jolli1 [7]

Answer:

33

Step-by-step explanation:

Since 31 + 32 + 33 + 34 = 130 and they are consecutive.

6 0
4 years ago
Read 2 more answers
A horticulturist working for a large plant nursery is conducting experiments on the growth rate of a new shrub. Based on previou
Jet001 [13]

Answer:

Test statistic = - 3.354

We reject H0 and conclude that the growth rate of the new shrub is less than 3cm per week.

Step-by-step explanation:

H0 : μ = 3

H1 : μ < 3

The test statistic = (xbar - μ) / (s/ √n)

Xbar = 1.50

μ = 3

s = 3

n = 45

Test statistic = (1.5 - 3) / (3/ √45)

Test statistic = - 1.5 / 0.4472135

Test statistic = - 3.354

Pvalue, using the Pvalue from Z score calculator :

Pvalue = 0.00039826

α = 0.05

Pvalue < α ; We reject H0 and conclude that the growth rate of the new shrub is less than 3cm per week.

8 0
3 years ago
Devin is collecting signatures for a petition to open a new park in her town. She needs to collect at least 1,000 signatures bef
steposvetlana [31]

Answer:

The correct opion is;

7 ≤ p

Step-by-step explanation:

The number of signatures Devin needs to collect = 1,000 signatures

The numer of signatures Devin has already = 380 signatures

The number ofsignatures each petition page can hold = 80 signatures

Therefore;

The number of extra signatures required = 1,000 - 380 = 680 signatures

The number of petition pages required = 1,000 petitions /(80 signatures/page)   = 12.5 pages

The number of pages already filled = 380 petitions /(80 signatures/page) = 4.75 pages

The number of pages remaining, p = 12.5 - 4.75 = 7.75 pages

Therefore, 7 ≤ p.

8 0
4 years ago
Read 2 more answers
Can someone explain me this please
AfilCa [17]

\purple{\maltese}\large\underline{\underline{\red{\sf\:\: Given :}}} \\

  • Mass (M) of space shuttle = 1.0 × 105 kg.
  • Altitude (r) = 200.0 km

\\\purple{\maltese}\large\underline{\underline{\red{\sf\:\: To \: \:   Find  :}}} \\

  • The force of gravity that the space shuttle experiences = ?

\\\purple{\maltese}\large\underline{\underline{\red{\sf\:\: Solution :}}} \\ \\

\purple{\maltese}\large\underline{\underline{\sf\:\: using \: formula :}} \\ \\

\bigstar \: \underline{ \boxed{\sf { \pink{g = \frac{G \: M}{{r }^{2} }}}}} \\  \\

\\ ❒\:  \: \underline{\textbf {Putting values \: in \: the \: \: given \: formula :}} \\

\sf \implies \: g = \frac{G \: M}{{r }^{2} }  \\

\sf \implies \: g = \frac{6.67×10^{-11} × 1.0×10^5}{(2×10^5)^{2} }  \\

\sf \implies \: g = \frac{6.67×10^{-6}}{4×10^{10}}\\

\sf \implies \: g = \frac{6.67×10^{-6}}{4×10^{10}} \\

\sf \implies \: g =1.6675×10^{-6-10}\\

\sf \implies \: g =1.6675×10^{-16}\: Newton\\\\

\:\:\:\:\:\:\:\qquad\rule{150pt}{2pt} \\\\

Henceforth, the force of gravity experienced by the shuttle is <u>1.6675×10^N</u>

4 0
2 years ago
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