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Marizza181 [45]
3 years ago
12

Help please ...................

Mathematics
2 answers:
ElenaW [278]3 years ago
7 0
94m squared . I just worked out the problem for you instead of doing my own homework lol
GarryVolchara [31]3 years ago
6 0

Answer:

2( \frac{1}{2}5 \times 6.4)  + 2( \frac{1}{2} 5 \times 12.4) \\  = 2(16) + 2(31) \\  = 32 + 62 \\  = 94m ^{2}

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I need help getting this answer right hurry
Rus_ich [418]

Answer:

Bring over -1 to get m = 10

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
When a scientist conducted a genetics experiments with peas, one sample of offspring consisted of 943 peas, with 717 of them hav
pshichka [43]

Using the normal approximation to the binomial distribution, it is found that:

a) 0.242 = 24.2% probability of getting 717 or more peas with red flowers.

b) Since Z < 2, 717 peas with red flowers is not significantly high.

c) Since 717 peas with red flowers is not a significantly high result, we cannot conclude that the scientist's assumption is wrong.

For each pea, there are only two possible outcomes. Either they have a red flower, or they do not. The probability of a pea having a red flower is independent of any other pea, which means that the binomial distribution is used to solve this question.

Binomial distribution:

Probability of x successes on n trials, with p probability.

Normal distribution:

In a normal distribution with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

  • It measures how many standard deviations the measure is from the mean.  
  • After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
  • If Z > 2, the result is considered <u>significantly high</u>.

If np \geq 10 and n(1-p) \geq 10, the binomial distribution can be approximated to the normal with:

\mu = np

\sigma = \sqrt{np(1-p)}

In this problem:

  • 943 peas, thus, n = 943
  • 3/4 probability of being red, thus p = \frac{3}{4} = 0.75.

Applying the approximation:

\mu = np = 943(0.75) = 707.25

\sigma = \sqrt{np(1-p)} = \sqrt{943(0.75)(0.25)} = 13.297

Item a:

Using continuity correction, this probability is P(X \geq 717 - 0.5) = P(X \geq 716.5), which is <u>1 subtracted by the p-value of Z when X = 716.5</u>.

Then:

Z = \frac{X - \mu}{\sigma}

Z = \frac{716.5 - 707.25}{13.297}

Z = 0.7

Z = 0.7 has a p-value of 0.758.

1 - 0.758 = 0.242

0.242 = 24.2% probability of getting 717 or more peas with red flowers.

Item b:

Since Z < 2, 717 peas with red flowers is not significantly high.

Item c:

Since 717 peas with red flowers is not a significantly high result, we cannot conclude that the scientist's assumption is wrong.

A similar problem is given at brainly.com/question/25212369

6 0
3 years ago
Please help
tigry1 [53]
Just plug in the x-values into the given equation to find. For 9-19-19 whatever the number is correlated to how much the graph is being moved down or up and the x is your slope.

5 0
3 years ago
PLEASE HELP
sweet-ann [11.9K]

Answer:

m < ANM = 36 degrees.

AM = 9.40 cm to the nearest hundredth.

Perimeter =  94.05 cm to the nearest hundredth.

Step-by-step explanation:

As we have a regular pentagon:

m < ANB = 360 / 5

= 72 degrees

So m < ANM = 1/2 * 72

= 36 degrees.

In the triangle ANM, AN = 16 so

sin 36 = Am / 16

AM =  16 sin36

= 9.4045 cm.

AB = 2 * AM = 18.809 cm

So as all sides are equal:

Perimeter = 5 * 18.809

= 94.05 cm.

4 0
3 years ago
The larger the standard deviation is for a data set, the more variable the data in the set are. True or false?
erik [133]
True.

If the standard deviation value is larger, it means the data points are further away from the mean - which indicates that the data in the set are more variable. 
3 0
3 years ago
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