The integer game, you have 3 cards. These cards add up to positive 6 and multiply to negative 64. What are the three cards in your hand?

Answers

Answer 1

Answer: mzybd 2 3 and 1 but idk lol

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-2x-6+(x/4)-3+90=180. Solve for x and show work.

Answers

so you want to add lik terms first
group them together
I have gropued the like terms using brackets
[-2x+(x/4)]+[-6-3+90]=180
add lik terms
add the number terms
-6-3+90=81
[-2x+(x/4)]+[81]=180
subtract 81 from both sides
-2x+(x/4)=99
add 2x to both sides
x/4=99+2x
multipliy both sides by 4 to clear fraction
x=396+8x
subtract 8x from both sides
-7x=396
mulitply both sides by -1
7x=-396
divide both sides by 7
x=-396/7=-56 and 4/7

If 3 pieces of candy cost $2.10,how much would four pieces cost?

Answers

We need to find the unit rate

2.10/3 = 0.70

0.70*4 = 2.80

Answer: 4 pieces would cost $2.80

210 divided by 3 equals 70

210+70= 280

$2.80

A survey was conducted to determine the average age at which college seniors hope to retire in a simple random sample of 101 seniors, 55 was theaverage desired retirement age, with a standard deviation of 3.4 years. A 96% confidence interval for desired retirement age of all college students is:
54.30 to 55.70
54.55 to 55.45
54.58 to 55.42
54 60 to 55.40

Answers

Answer:

96% confidence interval for desired retirement age of all college students is [54.30 , 55.70].

Step-by-step explanation:

We are given that a survey was conducted to determine the average age at which college seniors hope to retire in a simple random sample of 101 seniors, 55 was the average desired retirement age, with a standard deviation of 3.4 years.

Firstly, the Pivotal quantity for 96% confidence interval for the population mean is given by;

P.Q. = $(\bar X-\mu)/((s)/(√(n) ) )$ ~ $t_n_-_1$

where, $\bar X$ = sample average desired retirement age = 55 years

$\sigma$ = sample standard deviation = 3.4 years

n = sample of seniors = 101

$\mu$ = true mean retirement age of all college students

Here for constructing 96% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 96% confidence interval for the population mean, $\mu$ is ;

P(-2.114 < $t_1_0_0$ < 2.114) = 0.96 {As the critical value of t at 100 degree

of freedom are -2.114 & 2.114 with P = 2%}

P(-2.114 < $(\bar X-\mu)/((s)/(√(n) ) )$ < 2.114) = 0.96

P( $-2.114 * {(s)/(√(n) ) }$ < ${\bar X-\mu}$ < $2.114 * {(s)/(√(n) ) }$ ) = 0.96

P( $\bar X-2.114 * {(s)/(√(n) ) }$ < $\mu$ < $\bar X+2.114 * {(s)/(√(n) ) }$ ) = 0.96

96% confidence interval for $\mu$ = [ $\bar X-2.114 * {(s)/(√(n) ) }$ , $\bar X+2.114 * {(s)/(√(n) ) }$ ]

= [ $55-2.114 * {(3.4)/(√(101) ) }$ , $55+2.114 * {(3.4)/(√(101) ) }$ ]

= [54.30 , 55.70]

Therefore, 96% confidence interval for desired retirement age of all college students is [54.30 , 55.70].

Explain your reasoning as you compare the values of a^n and a^-n when n<0 brainly

Answers

n < 0, is another way to say "n is negative", so let's check

$\bf ~~~~~~~~~~~~\textit{negative exponents}\n\na^(-n) \implies \cfrac{1}{a^n}\qquad \qquad\cfrac{1}{a^n}\implies a^(-n)\qquad \qquada^n\implies \cfrac{1}{a^(-n)}\n\n[-0.35em]\rule{34em}{0.25pt}\n\na^n~\hspace{10.5em}\stackrel{n = -n}{a^(-n)}\implies \cfrac{1}{a^n}\n\n\na^(-n)~\hspace{10em}\stackrel{n=-n}{a^(-(-n))}\implies a^(+n)\implies a^n$