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3 years ago

PLEASE ANSWER ASSP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

2 answers:

miskamm [114]3 years ago

7 0

Answer: 1626815

Step-by-step explanation:

kow [346]3 years ago

5 0

Answer:

264 cm^3

Step-by-step explanation:

ASAP

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Jackie made $28 babysitting last week. her brother joe made only 86% as much as she did. how much did joe make?

strojnjashka [21]

Her brother joe made $52.08. First you need to make the percent into a decimal, like this 86% => .86. Then multiply .86 to $28. After that add what you got to $28.

7 0

3 years ago

In March, Audrey planted a tomato plant that was 11.25 inches tall. It grew 7.6 inches taller by July. How tall was the tomato s

m_a_m_a [10]

18.85 inches tall I believe. Tell me if am wrong

3 0

3 years ago

Please help solve with working much appreciated!

Lapatulllka [165]

The right answer is Option A.

Step-by-step explanation:

Given,

Heart rate goal = I

Heart rate reserve = H

70% of H = $\frac{70}{100}H = 0.7H$

Resting heart rate = R

According to given statement,

70% of heart rate reserved is added to resting heart rate to calculate heart rate goal, therefore,

I = R + 0.7H

I=R+0.7H represents the given situation.

The right answer is Option A.

Keywords: linear expression, addition

Learn more about addition at:

#LearnwithBrainly

8 0

3 years ago

Use the figure to answer the question that follows:

Olegator [25]

Answer:

m∠SQV + m∠VQT = 180°

Step-by-step explanation:

<u> m∠SQT is substituted by 180° at step 5</u>

4 m∠SQV + m∠VQT = m∠SQT Angle Addition Postulate
5 m∠SQV + m∠VQT = 180° Substitution Property of Equality
6 m∠VQT + m∠ZRS = 180° Same-Side Interior Angles Theorem

3 0

4 years ago

Suppose that, after measuring the duration of many telephone calls, a telephone company found their data was well-approximated b

Musya8 [376]

Answer:

a) 7.79%

b) 67.03%

c) Cumulative Distribution Function

$P(t) = \displaystyle\int^{\infty}_{-\infty} 0.1e^{-0.1t}~dt\\\\= \displaystyle\int^{b}_{a} 0.1e^{-0.1t}~dt, ~~a\leq t \leq b$

Step-by-step explanation:

We are given the following in the question:

$p(x) = 0.1 e^{-0.1x}$

where x is the duration of a call, in minutes.

a) P( calls last between 2 and 3 minutes)

$=\displaystyle\int^3_2 p(x)~ dx\\\\= \displaystyle\int^3_20.1e^{-0.1x}~dx\\\\=\Big[-e^{-0.1x}\Big]^3_2\\\\=-\Big[e^{-0.3}-e^{-0.2}\Big]\\\\= 0.0779\\=7.79\%$

b) P(calls last 4 minutes or more)

$=\displaystyle\int^{\infty}_4 p(x)~ dx\\\\= \displaystyle\int^{\infty}_40.1e^{-0.1x}~dx\\\\=\Big[-e^{-0.1x}\Big]^{\infty}_4\\\\=-\Big[e^{\infty}-e^{-0.4}\Big]\\\\=-(0- 0.6703)\\= 0.6703\\=67.03\%$

c) cumulative distribution function

$P(t) = \displaystyle\int^{\infty}_{-\infty} 0.1e^{-0.1t}~dt\\\\= \displaystyle\int^{b}_{a} 0.1e^{-0.1t}~dt, ~~a\leq t \leq b$

6 0

4 years ago