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2 years ago

Factorise 2-y(7-5y) please

1 answer:

hjlf2 years ago

5 0

Answer:
(y - 1)(5y - 2)

Solution:

Factor:
2 - y(7 - 5y)

Apply the Distributive Property:
2 - 7y + 5y²

Reorder the expression:
5y² - 7y + 2

Rewrite the term:
5y² - 5y - 2y + 2

Regroup terms into two proportional parts:
(5y² - 5y) + ( -2y + 2)

Factor Greatest Common Factors out:
5y(y - 1) - 2(y - 1)
(y - 1)(5y - 2)

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What are the measures of angle B and D

insens350 [35]

Answer:

Step-by-step explanation:⇔

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are the measures but if you forgot to add the picture then... who knows

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3 years ago

If you answer my question can you please tell me how to do it step by step.

hichkok12 [17]

Answer:

Depends

Step-by-step explanation:

It depends because I don't know if I'm strong on the material you're asking about.

5 0

3 years ago

Write as a decimal 23/9

Slav-nsk [51]

Answer:

2.555555555555.....

(the 5 is repeating)

Step-by-step explanation:

This is the answer because:

1) To convert the fraction to a decimal, you have to divide the numerator by the denominator.

2) Therefore, 23/9 is 2.5 (the 5 is repeating)

Hope this helps!

8 0

3 years ago

Line AC is tangent to circle O at point C<br> What is the measure of OAC?

Nuetrik [128]

AC is a tangent so by definition, it touches the circle at exactly one point (point C) and forms a right angle at the tangency point. So angle ACO is 90 degrees

The remaining angle OAC must be 45 degrees because we need to have all three angles add to 180
45+45+90 = 90+90 = 180

Alternatively you can solve algebraically like so
(angle OAC) + (angle OCA) + (angle COA) = 180
(angle OAC) + (90 degrees) + (45 degrees) = 180
(angle OAC) + 90+45 = 180
(angle OAC) + 135 = 180
(angle OAC) + 135 - 135 = 180 - 135
angle OAC = 45 degrees

Side Note: Triangle OCA is an isosceles right triangle. It is of the template 45-45-90.

7 0

3 years ago

A survey of 76 commercial airline flights of under 2 hours resulted in a sample average late time for a flight of 2.55 minutes.

nika2105 [10]

Answer:

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

Step-by-step explanation:

Information given

$\bar X=2.55$ represent the sample mean for the late time for a flight

$\mu$ population mean

$\sigma=12$ represent the population deviation

n=76 represent the sample size

Confidence interval

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

The confidence interval for the true mean is given by:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The Confidence level given is 0.95 or 95%, th significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ . If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got $z_{\alpha/2}=1.96$

Replacing we got:

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

7 0

3 years ago