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patriot [66]
2 years ago
13

On the moon, a bag of sugar has a weight of 3.7 Newtons (N) and a mass of 2.26 kilograms (kg). Which of the following describes

the mass of the sugar on Earth? A. more than its mass on the Moon B. same as its weight on the Moon. C. same as its mass on the Moon D. less than its mass on the Moon
Mathematics
1 answer:
ludmilkaskok [199]2 years ago
8 0

Answer:

c

Step-by-step explanation:

mass will never change only weight will as weight is dependent on gravitational field strength

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What is the perimeter of the smaller pentagon
Mice21 [21]

5 cm since a pentagon has 5 sides just multiply one side by 5

1x5=5

6 0
3 years ago
Identify whether the equation represents an exponential growth or exponential decay function.1. y = 1/4 (1/e)^-2x2. y = (1/e)^4x
zvonat [6]

The general formula for exponential growth and decays is:

y=y_0e^{kx}

if k>0 then then it is an exponential growth function. If k<0 then the function represents an exponential decay.

Now we need to classify each of the functions:

1.

The function

y=\frac{1}{4}(\frac{1}{e})^{-2x}

can be wrtten as:

\begin{gathered} y=\frac{1}{4}(e^{-1})^{-2x}^{} \\ =\frac{1}{4}e^{2x} \end{gathered}

comparing with the general formula we notice that k=2, therefore this is an exponential growth.

2.

The function

y=(\frac{1}{e})^{4x}

can be written as:

\begin{gathered} y=(\frac{1}{e})^{4x} \\ y=(e^{-1})^{4x} \\ y=e^{-4x} \end{gathered}

comparing with the general formula we notice that k=-4, therefore this is an exponential decay.

3.

The function

y=2e^{-x}+1

comparing with the general formula we notice that k=-1, therefore this is an exponential decay.

5 0
1 year ago
In 2008, data from the Center for Disease Control revealed that 28.5% of all male teenagers, aged 18-19 and attending U.S. colle
m_a_m_a [10]

Answer:

a) H0 : u = 28.5%

H1 : u < 28.5%

b) critical value = - 1.645

c) test statistic Z= - 1.41

d) Fail to reject H0

e) There is not enough evidence to support the professor's claim.

Step-by-step explanation:

Given:

P = 28.5% ≈ 0.285

X = 210

n = 800

p' = \frac{X}{n} = \frac{210}{800} = 0.2625

Level of significance = 0.05

a) The null and alternative hypotheses are:

H0 : u = 28.5%

H1 : u < 28.5%

b) Given a 0.05 significance level.

This is a left tailed test.

The critical value =

-Z_0.05 = -1.645

The critical value = -1.645

c) Calculating the test statistic, we have:

Z = \frac{p' - P}{\sqrt{\frac{P(1-P)}{n}}}

Z = \frac{0.2625 - 0.285}{\sqrt{\frac{28.5(1-28.5)}{800}}}

Z = -1.41

d) Decision:

We fail to reject null hypothesis H0, since Z = -1.41 is not in the rejection region, <1.645

e) There is not enough evidence to support the professor's claim that the proportion of obese male teenagers decreased.

5 0
3 years ago
Please help due today
Masja [62]

Answer: 80 ft²

Step-by-step explanation:

The area of a trapezoid is one half  the product of the height and the sum of the lengths of the  bases.

(6 + 10) * 0.5 * 10 = 80 (ft²)

3 0
3 years ago
A notebook costs 4.95 and the sales tax is 9.5 how much is the sales tax
bonufazy [111]

Answer:

$0.47

Step-by-step explanation:

8 0
3 years ago
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