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docker41 [41]
3 years ago
7

What is the slope of a line parallel to y =4x - 6?

Mathematics
2 answers:
BARSIC [14]3 years ago
7 0

Answer:

y = 5x - 6

Step-by-step explanation:

iragen [17]3 years ago
7 0
The slope would be -4.
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Which of the following is a binomial with degree 4?
wel

Answer:

the answer is A

Step-by-step explanation:

hope it's help

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3 years ago
What set of reflections would carry parallelogram ABCD onto itself?
My name is Ann [436]
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5 0
3 years ago
Read 2 more answers
Help me with the question in the pic please and thank u!
Zepler [3.9K]

Answer:

1.98

Step-by-step explanation:

In this problem, we are:

Given: ∠D = 26°, Hypotenuse (DF) = 4.5

To solve: EF

Now through the given, we are to focus on point D to solve for EF.

We need to find the relationship of point D to DF and EF.

The relationship here is sine.

Reason: \frac{opposite}{hypotenuse}

Through point D, DF is the hypotenuse and EF is the opposite side.

So let's solve this simply through algebra,

sin = \frac{opposite}{hypotenuse}

sin(26°) = \frac{EF}{4.5}

0.44 = \frac{EF}{4.5}

0.44 × 4.5 = EF

1.98 = EF

7 0
3 years ago
Write an equation in slope intercept form of a line passing through the given point (8,5) m=3
Zepler [3.9K]

If m(slope)=3, you can start to set up the equation as y=3x+b.

Using y=3x+8, you can plug in the given point being (8,5).

y=5

x=8

5=(3*8)+b

5=24+b

b=-19

The equation would be y=3x-19

3 0
3 years ago
he amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and s
sp2606 [1]

Complete question:

He amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and standard deviation 1.4 minutes. Suppose that a random sample of n equals 47 customers is observed. Find the probability that the average time waiting in line for these customers is

a) less than 8 minutes

b) between 8 and 9 minutes

c) less than 7.5 minutes

Answer:

a) 0.0708

b) 0.9291

c) 0.0000

Step-by-step explanation:

Given:

n = 47

u = 8.3 mins

s.d = 1.4 mins

a) Less than 8 minutes:

P(X

P(X' < 8) = P(Z< - 1.47)

Using the normal distribution table:

NORMSDIST(-1.47)

= 0.0708

b) between 8 and 9 minutes:

P(8< X' <9) =[\frac{8-8.3}{1.4/ \sqrt{47}}< \frac{X'-u}{s.d/ \sqrt{n}} < \frac{9-8.3}{1.4/ \sqrt{47}}]

= P(-1.47 <Z< 6.366)

= P( Z< 6.366) - P(Z< -1.47)

Using normal distribution table,

NORMSDIST(6.366)-NORMSDIST(-1.47)

0.9999 - 0.0708

= 0.9291

c) Less than 7.5 minutes:

P(X'<7.5) = P [Z< \frac{7.5-8.3}{1.4/ \sqrt{47}}]

P(X' < 7.5) = P(Z< -3.92)

NORMSDIST (-3.92)

= 0.0000

3 0
3 years ago
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