2.5(25) +75<br><br><br> Haha I don’t really understand this!

The Intelligence Quotient (IQ) test scores for adults are normally distributed with a population mean of 100 and a population st

adell [148]

Answer: 0.7471

Step-by-step explanation:

Given : The Intelligence Quotient (IQ) test scores for adults are normally distributed with a population mean of $\mu=100$ and a population standard deviation of $\sigma=15$ .

Sample size = 50

Using formula $z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}$ , the z-value corresponding to x=98 will be:-

$z=\dfrac{98-100}{\dfrac{15}{\sqrt{50}}}\approx-0.94$

Z-value corresponding to x=103 will be:-

$z=\dfrac{103-100}{\dfrac{15}{\sqrt{50}}}\approx1.41$

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

P-value = $P(98$

Hence, the probability we could select a sample of 50 adults and find that the mean of this sample is between 98 and 103 = 0.7471

8 0

4 years ago

Can you help me with question 6 please

Licemer1 [7]

I'm pretty sure the answer is 24

4 0

3 years ago

Can someone help me with this math question. Will mark Brainliest. I posted a picture of the question.

Romashka-Z-Leto [24]

Answer:

Step-by-step explanation:

∠2= ∠4 { alternate interior angles}

∠2= 65

∠1 + ∠2 = 180 {linear pair}

∠1 + 65 = 180

∠1 = 180- 65

∠1 = 115

4 0

3 years ago

NO LINKS!!! Find the arc measure and arc length of AB. Then find the area of the sector ABQ.

Norma-Jean [14]

Answer:

<u>Arc Measure</u>: equal to the measure of its corresponding central angle.

<u>Formulas</u>

$\textsf{Arc length}=2 \pi r\left(\dfrac{\theta}{360^{\circ}}\right)$

$\textsf{Area of a sector of a circle}=\left(\dfrac{\theta}{360^{\circ}}\right) \pi r^2$

$\textsf{(where r is the radius and the angle }\theta \textsf{ is measured in degrees)}$

<h3><u>Question 39</u></h3>

Given:

r = 7 in
$\theta$ = 90°

Substitute the given values into the formulas:

Arc AB = 90°

$\textsf{Arc length of AB}=2 \pi (7) \left(\dfrac{90^{\circ}}{360^{\circ}}\right)=3.5 \pi=11.00\:\sf in\:(2\:d.p.)$

$\textsf{Area of the sector AQB}=\left(\dfrac{90^{\circ}}{360^{\circ}}\right) \pi (7)^2=\dfrac{49}{4} \pi=38.48\:\sf in^2\:(2\:d.p.)$

<h3><u>Question 40</u></h3>

Given:

r = 6 ft
$\theta$ = 120°

Substitute the given values into the formulas:

Arc AB = 120°

$\textsf{Arc length of AB}=2 \pi (6) \left(\dfrac{120^{\circ}}{360^{\circ}}\right)=4\pi=12.57\:\sf ft\:(2\:d.p.)$

$\textsf{Area of the sector AQB}=\left(\dfrac{120^{\circ}}{360^{\circ}}\right) \pi (6)^2=12 \pi=37.70\:\sf ft^2\:(2\:d.p.)$

<h3><u>Question 41</u></h3>

Given:

r = 12 cm
$\theta$ = 45°

Substitute the given values into the formulas:

Arc AB = 45°

$\textsf{Arc length of AB}=2 \pi (12) \left(\dfrac{45^{\circ}}{360^{\circ}}\right)=3 \pi=9.42\:\sf cm\:(2\:d.p.)$

$\textsf{Area of the sector AQB}=\left(\dfrac{45^{\circ}}{360^{\circ}}\right) \pi (12)^2=18 \pi=56.55\:\sf cm^2\:(2\:d.p.)$

8 0

2 years ago

Let f(x) = x2 +6x +8 . What is the domain of Kx)?

Vedmedyk [2.9K]

Answer:

The domain is all real numbers

Step-by-step explanation:

f(x) = x^2 +6x +8

The domain is what numbers can we put in for x

This is a parabola, we can put in any number for x and the function is defined

8 0

3 years ago

2.5(25) +75 Haha I don’t really understand this!