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ratelena [41]
3 years ago
5

(10) Jasmine uses 14.24 pounds of fruit for

Mathematics
1 answer:
solniwko [45]3 years ago
5 0
The answer is 0.89 pounds of fruit per serving- please give brainliest answer!
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Identify the triangles that are similar by the AA similarity theorem. explain how you know that the triangles are similar.
UNO [17]

Angle WVX = Angle YVZ (vertically opposite angles)

Angle VWX = Angle VYZ (shown)

Angle WXV = Angle YZV (total sum of angles in triangle)

4 0
2 years ago
Which expression can be used to find the value of the expression below -4(5 + (-2)​
guapka [62]

Answer:

-12

Step-by-step explanation:

-4(5+(-2))=-4(5-2)=-4(3)=-12

6 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
Cassie has 1/2 pound of sugar in her cabinet.Her cake recipe calls
BigorU [14]

Answer:

2 cakes with the sugar she has and there will still be some sugar left.

Step-by-step explanation:

Amount of sugar in the cabinet of Cassie = 1/2 pound

Amount of sugar required by Cassie to make her cake recipe = 2/10 pound

                                                                                                  = 1/5 pounds

Then

The number of cakes that Cassie can make = (1/2)/(1/5)

                                                                       = 5/2

                                                                       = 2 1/2




6 0
3 years ago
Last spring 12 birds ate 8 cups of birdseed in one month. If I am expecting 18 birds this month, how
elena55 [62]

Simplified Ratio:  <u>3 : 2</u>

Birds / Cups

6 / 4

9 / 6

12 / 8

15 / 10

18 / 12

You need 12 cups of birdseed if you're expecting 18 birds this month.

7 0
3 years ago
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