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3 years ago

I think the answer is A but I'm not sure! Can someone help me? Thank you!

Mathematics

2 answers:

babunello [35]3 years ago

3 0

The answer is A cause if you solve the problem it will be x < 10 and A is going to the left which is less than 10 so the answer is A.

ElenaW [278]3 years ago

3 0

I think you are right.

4x + 20 < 40 Subtract 20 from both sides

4x < 40 - 20

4x < 20 Divide by 4

x < 20 / 4

x < 5 Somewhere very near 5 is as close as the solution gets. So at 5 you use an open circle. Because near 5 is the largest possible solution, the arrow heads left. Pat yourself on the back. These things are not easy.

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8(-7+n)=-128 please help

kow [346]

N = -9
8(-7+n) = -128
Divide both sides by 8
-7 + n = -16
Add 7 to both sides
n = -9

8 0

3 years ago

ASAP!!! please help! thanks.

Neko [114]

Answer:

I think A is the right answer

4 0

3 years ago

Solve the equation for a.

Pavel [41]

Answer:

$\large\boxed{a=\dfrac{K}{4+9b}}$

Step-by-step explanation:

$K=4a+9ab\qquad\text{use the distributive property}\ a(b+c)=ab+ac\\\\K=a(4+9b)\qquad\text{divide both sides by}\ (4+9b)\neq0\\\\\dfrac{K}{(4+9b)}=\dfrac{a(4+9b)}{(4+9b)}\\\\\dfrac{K}{4+9b}=a$

6 0

3 years ago

Read 2 more answers

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

a boat capsized and sank in the lake. based on an assumption of a mean weight of 143 lb, the boat was rated to carry 70 passenge

Alenkinab [10]

The final answer is the probability that the boat is overloaded because the mean weight of the 70 passengers is greater than 143lb is 1.0000.

A normal distribution is a continuous data distribution with a bell-shaped curve. The normally distributed random variable X has mean $\mu$ and standard deviation $\sigma$ .

In addition, the standard normal distribution represents a normal curve with a mean of zero and a standard deviation of one.

The number of standard deviations an element deviates from the mean is indicated by a standardized Z-score.

What is the theorem of the central limit?

If the sample size from the population is sufficiently large and has a finite variance, the mean of all samples taken will be approximately the same as the population mean. And the variance would be the population variance divided by the sample size. In addition, the sampling distribution would be roughly normal.
The formula for sampling distribution of the sample mean is $\mu_{\bar{x}}=\mu$
The formula for standard deviation of the sample mean is, $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$
Here, $\sigma$ is the population standard deviation and n is the sample size.
The formula for standard z-score is,

$z=\frac{\bar{x}-\mu_{x}}{\frac{\sigma}{\sqrt{n}}}$

From the given information,Population mean, $\mu=171$ ,Population standard deviation, $\sigma=37.7$ ,Sample size, n=70
Compute the probability that the boat is overloaded because the mean weight of the 70 passengers is greater than $143l \mathrm{~b}$ .

$\begin{aligned}P(\bar{X} > 143) &=P\left(\frac{\bar{X}-\mu}{\sigma / \sqrt{n}} > \frac{143-171}{37.7 / \sqrt{70}}\right) \\&=P(Z > -6.21) \\&=1-P(Z < -6.21) \text { (Use the Excel function NORMSDIST(-6.21)) } \\&=1-0.0000 \\&=1.0000\end{aligned}$

To learn more about standard deviation visit:

brainly.com/question/16555520?

#SP4

4 0

2 years ago