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3 years ago

Need help anything helps thanks

Mathematics

1 answer:

yKpoI14uk [10]3 years ago

5 0

A binomial squared is written like

$x^2+2ax+a^2$

So, we have $2a=6 \iff a=3$

So, we want to complete the square $(x+3)^2 = x^2+6x+9$

We can add and subtract 9 to write

$x^2+6x = x^2+6x+(9-9) = (x^2+6x+9)-9 = (x+3)^2-9$

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Consider the one-sided confidence interval expressions for the mean of a normal population. Round your answers to 2 decimal plac

ser-zykov [4K]

Answer: a) 0.84 b) 0.67 c) 1.28

Step-by-step explanation:

Using the standard normal distribution table for z-value , we have

(a) The value of $z_{\alpha}$ would result in a 80% one-sided confidence interval : $z_{(1-0.80)}=z_{0.20}=0.8416\approx0.84$

(b) The value of $z_{\alpha}$ would result in a 85% one-sided confidence interval : $z_{(1-0.85)}=z_{0.25}=0.6744897\approx0.67$

(c) The value of $z_{\alpha}$ would result in a 90% one-sided confidence interval : $z_{(1-0.90)}=z_{0.10}=1.2815515\approx1.28$

6 0

3 years ago

I really need help with this!!!

netineya [11]

2 because is isisisisisisisisis

5 0

3 years ago

3 1/2 • 4 3/4 please help

Dmitry [639]

Answer: 16 5/8

Step-by-step explanation:

3 1/2 = 7/2

4 3/4 = 19/4

7/2 * 19/4 = 133/8 or 16 5/8

3 0

3 years ago

Read 2 more answers

The price of an item has risen to $152 today. Yesterday it was $95. Find the percentage increase.

Oksanka [162]

Answer:

60% increase

Step-by-step explanation:

152 - 95 = -57

57 / 95 = 0.6

0.6 x 100 = 60

5 0

2 years ago

An aeroplane X whose average speed is 50°km/hr leaves kano airport at 7.00am and travels for 2 hours on a bearing 050°. It then

Zigmanuir [339]

Answer:

(a)123 km/hr

(b)39 degrees

Step-by-step explanation:

Plane X with an average speed of 50km/hr travels for 2 hours from P (Kano Airport) to point Q in the diagram.

Distance = Speed X Time

Therefore: PQ =50km/hr X 2 hr =100 km

It moves from Point Q at 9.00 am and arrives at the airstrip A by 11.30am.

Distance, QA=50km/hr X 2.5 hr =125 km

Using alternate angles in the diagram:

$\angle Q=110^\circ$

(a)First, we calculate the distance traveled, PA by plane Y.

Using Cosine rule

$q^2=p^2+a^2-2pa\cos Q\\q^2=100^2+125^2-2(100)(125)\cos 110^\circ\\q^2=34175.50\\q=184.87$ km$

SInce aeroplane Y leaves kano airport at 10.00am and arrives at 11.30am

Time taken =1.5 hour

Therefore:

Average Speed of Y

$=184.87 \div 1.5\\=123.25$ km/hr\\\approx 123$ km/hr (correct to three significant figures)$

(b)Flight Direction of Y

Using Law of Sines

$\dfrac{p}{\sin P} =\dfrac{q}{\sin Q}\\\dfrac{125}{\sin P} =\dfrac{184.87}{\sin 110}\\123 \times \sin P=125 \times \sin 110\\\sin P=(125 \times \sin 110) \div 184.87\\P=\arcsin [(125 \times \sin 110) \div 184.87]\\P=39^\circ $ (to the nearest degree)$

The direction of flight Y to the nearest degree is 39 degrees.

7 0

3 years ago