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insens350 [35]
3 years ago
14

Find the equation of the line passing through the given points. Use function notation to write the equation, (-2,8) and (-6,20)

Mathematics
1 answer:
sertanlavr [38]3 years ago
7 0

-6-(-2)=4

20-8=12

4/12=3

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Leela made goody bags for her party and put 5 pieces of candy in each bag. She bought gummies and chocolates yo put in het goody
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Answer:

4 ways

Step-by-step explanation:

1 gummy 4 chocolates

2 gummy 3 chocolates

3 gummy 2 chocolates

4 gummy 1 chocolates

This is what i think the answer is hope it helps

5 0
3 years ago
If Ali wears a green shirt then we will take a test if we take a test then everyone will pass the test
tresset_1 [31]

Answer:

yea only ofc

Step-by-step explanation:

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3 years ago
Based on the 2009 season, the Texas Rangers have a winning percentage of .533. Use the binomial model to find the probability th
Naddik [55]

Answer:

0.128

Step-by-step explanation:

We know the probability for any event X is given by,

P(X=x)=\binom{n}{x}\times p^{n-x}\times q^{x},

where p is the probability of success and q is the probability of failure.

Here, we are given that p = 0.533.

Since, we have that q = 1 - p

i.e. q = 1 - 0.533

i.e. q = 0.467

It is required to find the probability of 4 wins in the next 5 games i.e. P(X=4) when n = 5.

Substituting the values in the above formula, we get,

P(X=4)=\binom{5}{4}\times 0.533^{5-4}\times 0.467^{4}

i.e. P(X=4)=5 \times 0.533 \times 0.048

i.e. P(X=4)=5 \times 0.533 \times 0.048

i.e. i.e. P(X=4)=0.128

Hence, the probability of 4 wins in the next 5 games is 0.128.

8 0
3 years ago
Read 2 more answers
Which answer describes the simpler problems that could be used to solve this story problem?
puteri [66]
<span>B. Calculate how many pieces of fruit go into each bag. Multiply this by the number of students going on the trip. </span>
4 0
3 years ago
Is there any systematic tendency for part-time college faculty to hold their students to different standards than do full-time f
alexandr1967 [171]

Answer:

H0 : μ1 - μ2 = 0

H1 : μ1 - μ2 ≠ 0

-1. 34

0.1837

Step-by-step explanation:

Full time :

n1 = 125

x1 = 2.7386

s1 = 0.65342

Part time :

n2 = 88

x2 = 2.8439

s2 = 0.49241

H0 : μ1 - μ2 = 0

H1 : μ1 - μ2 ≠ 0

Test statistic :

The test statistic :

(x1 - x2) / sqrt[(s1²/n1 + s2²/n2)]

(2.7386 - 2.8439) / sqrt[(0.65342²/125 + 0.49241²/88)]

−0.1053 / sqrt(0.0034156615712 + 0.0027553)

-0.1053 /0.0785554

= - 1.34

Test statistic = - 1.34

The Pvalue :

Using df = smaller n - 1 = 88 - 1 = 87

Pvalue from test statistic score ;

Pvalue = 0.1837

Pvalue > α ; We fail to reject the null and conclude that the GPA does not differ.

At α = 0.01 ; the result is insignificant

6 0
2 years ago
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