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Doss [256]
3 years ago
14

Help me plz cant work it out

Mathematics
1 answer:
svetoff [14.1K]3 years ago
7 0

Answer: 27.80

Step-by-step explanation:

(3/5) x 14 = 8.4

That's 4 of the 2 kilogram tubs and 1 of the 1 kilogram tub (gives 9 kilograms)

4 x 6.15 = 24.60

1 x 3.20 = 3.20

24.60 + 3.20 = 27.80

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4 4/9 simplified into the smallest from
Rasek [7]
1/9 maybe possible go with the best choice
3 0
3 years ago
Assume that females have pulse rates that are normally distributed with a mean of μ=73.0 beats per minute and a standard deviati
Gennadij [26K]

Answer:

a. the probability that her pulse rate is less than 76 beats per minute is 0.5948

b. If 25 adult females are randomly​ selected,  the probability that they have pulse rates with a mean less than 76 beats per minute is 0.8849

c.   D. Since the original population has a normal​ distribution, the distribution of sample means is a normal distribution for any sample size.

Step-by-step explanation:

Given that:

Mean μ =73.0

Standard deviation σ =12.5

a. If 1 adult female is randomly​ selected, find the probability that her pulse rate is less than 76 beats per minute.

Let X represent the random variable that is normally distributed with a mean of 73.0 beats per minute and a standard deviation of 12.5 beats per minute.

Then : X \sim N ( μ = 73.0 , σ = 12.5)

The probability that her pulse rate is less than 76 beats per minute can be computed as:

P(X < 76) = P(\dfrac{X-\mu}{\sigma}< \dfrac{X-\mu}{\sigma})

P(X < 76) = P(\dfrac{76-\mu}{\sigma}< \dfrac{76-73}{12.5})

P(X < 76) = P(Z< \dfrac{3}{12.5})

P(X < 76) = P(Z< 0.24)

From the standard normal distribution tables,

P(X < 76) = 0.5948

Therefore , the probability that her pulse rate is less than 76 beats per minute is 0.5948

b.  If 25 adult females are randomly​ selected, find the probability that they have pulse rates with a mean less than 76 beats per minute.

now; we have a sample size n = 25

The probability can now be calculated as follows:

P(\overline X < 76) = P(\dfrac{\overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{ \overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}})

P( \overline X < 76) = P(\dfrac{76-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{76-73}{\dfrac{12.5}{\sqrt{25}}})

P( \overline X < 76) = P(Z< \dfrac{3}{\dfrac{12.5}{5}})

P( \overline X < 76) = P(Z< 1.2)

From the standard normal distribution tables,

P(\overline X < 76) = 0.8849

c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

In order to determine the probability in part (b);  the  normal distribution is perfect to be used here even when the sample size does not exceed 30.

Therefore option D is correct.

Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

5 0
2 years ago
In the rhombus m&lt;1=160. What are m&lt;2 and &lt;3? The diagram is not drawn to scale.
xxMikexx [17]
Since a rhomubs is a 4 sided quadrilateral when parallel lines intersect within we solve for the triangles that are the result of the intersecting lines m<2=10, m<3=10
8 0
3 years ago
A group of friends goes out to lunch. The total bill, excluding tax, is $22. If tax on the meal is 6% and if a 20% tip is left o
TiliK225 [7]

Answer:

27.72

Step-by-step explanation:

Sorry about before i made a slight mistake

5 0
1 year ago
A video game club charges a fixed annual membership fee of $20 and $2 per video game rented. Let f(n) represent the total annual
Rasek [7]
<span>fixed annual membership fee of $20
</span><span>$2 per video game rented.
</span><span>Let f(n) represent the total annual cost of renting n video games
so
f(n) = 20 + 2n

if </span><span>increased by $15 the next year
then
f(n) = </span>20 + 2n + 15
f(n) = 2n + 35

answer
<span>f(n) = 2n + 35 (first choice)</span>
5 0
3 years ago
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