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4 years ago

This is math please help

Mathematics

1 answer:

elena-s [515]4 years ago

6 0

-2x+6= 28
you want to get (2x) by itself so you subtract 6 from 28 on both sides
-2x= 22
then you want to get x by itself so you divide 2 on both sides
X= -11
Hope this helps!!! You could also use photomath to see if your answer is correct

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|x+6| =8 A) x=-2 and x= -14 B) x= -2 and x= 14 C) x=2 and x= -14 D) x=2 and x= -2

erica [24]

Answer:

C) x=2 and x= -14

Step-by-step explanation:

|x + 6| = 8

means

x + 6 = 8 or x + 6 = -8

x + 6 = 8

x = 2

x + 6 = -8

x = -14

5 0

3 years ago

In a survey, 28 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped

Rina8888 [55]

Answer:

The 80% confidence level estimate for how much a typical parent would spend on their child's birthday gift is between $33.954 and $35.246.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution should be used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 28 - 1 = 27

80% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 27 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.8}{2} = 0.9$ . So we have T = 1.314

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 1.314\frac{2.6}{\sqrt{28}} = 0.646$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 34.6 - 0.646 = $33.954

The upper end of the interval is the sample mean added to M. So it is 34.6 + 0.646 = $35.246

The 80% confidence level estimate for how much a typical parent would spend on their child's birthday gift is between $33.954 and $35.246.

3 0

3 years ago

Please help math questions put number with answers please

natita [175]

Answer:

See below for answers

Step-by-step explanation:

Problem 1

$\frac{1}{sin\theta}=2cos\theta\: ; \: 0\leq\theta < 2\pi\\\\1=2sin\theta cos\theta\\\\1=sin2\theta\\\\\frac{\pi}{2}+2\pi n=2\theta\\ \\\frac{\pi}{4}+\pi n=\theta\\ \\\theta=\bigr\{\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\bigr\}$

Therefore, the second option is correct.

Problem 2

$cos(2x)+1=sin(x)+2\:;\: 0\leq x < 2\pi\\\\1-2sin^2x+1=sin(x)+2\\\\-2sin^2x=sin(x)\\\\0=2sin^2x+sinx\\\\0=sinx(2sinx+1)$

$0=sinx\\\\x=\pi n\\\\x=0,\pi$

$0=2sinx+1\\\\-1=2sinx\\\\-\frac{1}{2}=sinx\\ \\x=\frac{7\pi}{6}+2\pi n,\frac{11\pi}{6}+2\pi n\\\\x=\frac{7\pi}{6},\frac{11\pi}{6}$

Therefore, the solution set is $x=\bigr\{0,\pi,\frac{7\pi}{6},\frac{11\pi}{6}\bigr\}$ , making the second option correct.

Problem 3

$2cos^2x=-cosx\: ;\: 0^\circ\leq x < 360^\circ\\\\2cos^2x+cosx=0\\\\cosx(2cosx+1)=0$

$cosx=0\\\\x=\frac{\pi}{2}+\pi n\\ \\x=90^\circ+180n^\circ\\\\x=90^\circ,270^\circ$

$2cosx+1=0\\\\2cosx=-1\\\\cosx=-\frac{1}{2}\\\\x=\frac{2\pi}{3}+2\pi n,\frac{4\pi}{3}+2\pi n\\ \\x=120^\circ+360n^\circ,240^\circ+360n^\circ\\\\x=120^\circ,240^\circ$