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Tanya [424]
3 years ago
11

The table below represents the relationship of the amount of snowfall

Mathematics
1 answer:
RUDIKE [14]3 years ago
7 0

Answer:

yes in 4 of the five countries the snowfall in inches is double the amount of time in hours

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Let E be the event that a corn crop has an infestation of ear worms, and let B be the event that a corn crop has an infestation
KatRina [158]

Answer:

The probability that a corn crop has either an ear worm infestation, a corn borer infestation

P(EUB) = 0.27

Step-by-step explanation:

<u>Explanation</u>:-

<u>Addition theorem on probability</u>:-

If S is a sample space, and E , F are any events in S then

P(EUF) = P(E) +P(F) -P(E n F)

Let 'E' be the event that a corn crop has an infestation of ear worms

let 'B' be the event that a corn crop has an infestation of corn bores

P(EUB) = P(E) +P(B) -P(E n B)

given P(E) = 0.24 and P(B) = 0.16 and P(E n B) =0.13

P(EUB) = P(E) +P(B) -P(E n B)

P(EUB) = 0.24 + 0.16 - 0.13

           = 0.27

The probability that a corn crop has either an ear worm infestation, a corn borer infestation

P(EUB)=0.27

7 0
4 years ago
32a^6b^5 <br> 26a^2b^4 <br> greatest common factor
irga5000 [103]

Answer:

2a^2b^4

Step-by-step explanation:

7 0
4 years ago
An author argued that more basketball players have birthdates in the months immediately following July​ 31, because that was the
Lady_Fox [76]

Answer:

There is sufficient evidence to warrant rejection of the claim that professional basketball players are born in different months with the same​ frequency.

Step-by-step explanation:

In this case we need to test whether there is sufficient evidence to warrant rejection of the claim that professional basketball players are born in different months with the same​ frequency.

A Chi-square test for goodness of fit will be used in this case.

The hypothesis can be defined as:

<em>H₀</em>: The observed frequencies are same as the expected frequencies.

<em>Hₐ</em>: The observed frequencies are not same as the expected frequencies.

The test statistic is given as follows:

 \chi^{2}=\sum{\frac{(O-E)^{2}}{E}}

The values are computed in the table.

The test statistic value is \chi^{2}=128.12.

The degrees of freedom of the test is:

n - 1 = 12 - 1 = 11

Compute the <em>p</em>-value of the test as follows:

p-value < 0.00001

*Use a Chi-square table.

p-value < 0.00001 < α = 0.05.

So, the null hypothesis will be rejected at any significance level.

Thus, there is sufficient evidence to warrant rejection of the claim that professional basketball players are born in different months with the same​ frequency.

6 0
3 years ago
David writes down the sequence 2, 6, 10, 14<br><br>he says the sequence is n+4 <br><br>is he correct
Sonja [21]
He is, since each number is 4 more in value than the previous number
7 0
3 years ago
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Find the value of X for the given parallelogram
horrorfan [7]
The value of X for the given parallelogram is 8
4 0
2 years ago
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