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3 years ago

Help Please... 15 points :)

Mathematics

2 answers:

morpeh [17]3 years ago

5 0

Answer:

√100 = 10, √36 = 6, √225 = 15

Step-by-step explanation:

100 can be made by 10^2 (10 x 10), 36 has factors of 6 and can be made by 6^2, and 225 can be made by 15^2.

liraira [26]3 years ago

3 0

$\sqrt{100} = \sqrt{10 \times 10}$ = $\boxed{10}$

$\sqrt{36} = \sqrt{6 \times 6}$ = $\boxed{6}$

$\sqrt{225} = \sqrt{15 \times 15}$ = $\boxed{15}$

$\red{\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese \: \: Mystique35}}}}}$

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10% of 65 pls help and explain the steps to get total answer

sergij07 [2.7K]

10% of 65
Of means multiply
=0.1 x 65
=6.5

Or you can use another way which is (part/whole) *(percent/100)

Whole=65
Percent=10
And part(part of 65)=unknown

(X/65)* (10/100)
Cross multiply
x= 6.5

So 10% of 65 is 6.5

5 0

2 years ago

The amount of polonium-210 remaining, p(t), after t days in a sample can be modeled by the exponential function p(t) = 100e−0.00

Archy [21]

Answer:

% Po lost = 100[1 - e^(-0.005t)] %; 73.0 g

Step-by-step explanation:

p(t) = 100e^(-0.005t)

Initial amount: p(0) = 100

Amount remaining: p(t) = 100e^(-0.005t)

Amount lost: p(0) – p(t) = 100 - 100e^(-0.005t) = 100[1 - e^(-0.005t)]

% of Po lost = amount lost/initial amount × 100 %

= [1 - e^(-0.005t)] × 100 % = 100[1 - e^(-0.005t)] %

p(63) = 100e^(-0.005 × 63) = 100e^(-0.315) = 100 × 0.730 = 73 g

The mass of polonium remaining after 63 days is 73 g.

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2 years ago

Consider the problem: After charging for 1/3 of an hour, a phone is at 2/5 of its full power. How long will it take the phone to

prisoha [69]

Answer:

Equation 2

is the answer

Step-by-step explanation:

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2 years ago

Read 2 more answers

A math instructor assigns a group project in each of the scenarios below, the instructor selects 5 consecutive students from thi

SVETLANKA909090 [29]

Complete question:

Consider a class of 20 students consisting of 5 sophomores, 8 juniors, and 7 seniors. A math instructor assigns a group project in each of the scenarios below, the instructor selects 5 consecutive students from this class, keeping track (in order) of the level of the student that he gets.

1a. How many possible outcomes are in the sample space S? (An outcome is a 5- tuple of students.)

Let A2 denote the event that exactly 2 of the selected students are sophomore, A5 denote the event that exactly 5 of the selected students are the juniors, and A4 denote the event that exactly 4 of the selected students are seniors.

1b. Find the probabilities of each of these three events. A2, A5, A4

1c. Do the events A2, A5, A4 constitute a partition of the sample space?

Answer:

Given:

n = 20

a) The possible outcome in the sample space,S, =

ⁿCₓ = ²⁰C₅

$= \frac{20!}{(20-5)! 5!)}$