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Liula [17]
3 years ago
13

Crystal and Victor are comparing the estimated calories burned in a calorie-counting software to the actual calories burned duri

ng exercise.
Use the drop-down menus to complete the statements about the software's accuracy.

Mathematics
1 answer:
prohojiy [21]3 years ago
4 0

Answer:

ok so im on the same problem here lets work this through

obviously do the first 4 boxes cuz those r easy then get help on the last 2

Step-by-step explanation:

You might be interested in
26. Students who take a statistics course are given a pre-test on the concepts and skills for the first chapter of a statistics
hjlf

The test statistic value lies to the right of the critical value. So we have sufficient evidence to reject the null hypothesis.

<h3>What are null hypotheses and alternative hypotheses?</h3>

In null hypotheses, there is no relationship between the two phenomenons under the assumption or it is not associated with the group. And in alternative hypotheses, there is a relationship between the two chosen unknowns.

Students who take a statistics course are given a pre-test on the concepts and skills for the first chapter of a statistics course.

Then they are given a post-test once the professor has concluded lecturing on the material.

Pre-test and post-test scores for 4 students in an elementary statistics class are given below.

Then we have

\mu _d = \mu _{post} - \mu _{pre}

Then the null hypotheses and alternative hypotheses will be

H₀: \mu _d = 0

Hₐ: \mu _d > 0

Then the test statistic will be

\rm \overline{x} _d = \dfrac{\Sigma x_d}{n} = \dfrac{15+12+10+1}{4}\\\\\overline{x} _d = 9.5

Then

\rm S_d = 6.02

The test statistic value is given by

\rm t = \dfrac{\overline{x} _d }{\dfrac{S_d}{\sqrtn}} \\\\t = \dfrac{9.5}{\dfrac{6.02}{\sqrt4}}\\\\t = 3.16

Since this is a right-tailed test, so the critical value is given by

\rm t_{n-1}(\alpha ) = t_3 (0.05) = 2.353

Since the test statistic value lies to the right of the critical value. So we have sufficient evidence to reject the null hypothesis.

Hence, we can conclude that \mu _d > 0 that is test scores have improved.

More about the null hypotheses and alternative hypotheses link is given below.

brainly.com/question/9504281

#SPJ1

7 0
2 years ago
A medication with an initial concentration of 1. 5 milligrams per liter is administered to a
lesya [120]

Answer: Exponential decay model:

y

=

x

(

1

−

r

)

t

, half life of tablet is about

2

hours and after

t

=

3

hours , remaining drug on patient's system is

42.875

mg.

Step-by-step explanation: Initial drug

x

=

125

mg ; rate of decay

r

=

30

100

=

0.3

gm/hour

Exponential model:

y

=

x

(

1

−

r

)

t

=

125

(

1

−

0.3

)

t

=

125

⋅

0.7

t

Half life:

y

=

125

2

=

62.5

mg

∴

62.5

=

125

⋅

0.7

t

or

0.7

t

=

1

2

. Taking logarithm on both sides we get ,

t

log

(

0.7

)

=

log

(

0.5

)

∴

t

=

log

(

0.5

)

log

(

0.7

)

≈

1.94

(

2

d

p

)

hour

The half life of tablet is about

2

hours.

After

t

=

3

hours , remaining drug on patient's system is

y

=

125

⋅

0.7

t

=

125

⋅

0.7

3

=

42.875

mg [Ans]

8 0
2 years ago
A diagonal support beam is attached to the gate of a fence as shown. Which angles must be congruent in order for the top and bot
ludmilkaskok [199]

Answer:

3 and 2

Step-by-step explanation:


4 0
3 years ago
Estimate the area of the circle.
mariarad [96]

Answer:

4

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Is number 1 true or false
vekshin1

Answer:

true

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
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