Solve for y 6x+5y=12

Mathematics

2 answers:

OLEGan [10]3 years ago

4 0

I’m pretty sure this is what you do for these

krek1111 [17]3 years ago

3 0

I got 0.4, not sure if that’s correct, but i solved it as if there was only one variable. since it’s 5*y i did the opposite and divided and got 6x+y=2.4. then i did 6/6 and did 2.4/6 and got x+y=0.4. i’d wait till more people answer to be sure that is correct

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Pauley graphs the change in temperature of a glass of hot tea over time. He sees that the function appears to decrease quickly a

denpristay [2]

As the liquid gets closer to the same temperature as the room, it cools down slower. If you put an glass of hot cocoa in a really cold room, it'll decrease really fast until it's around the same temperature as the room

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3 years ago

A farmer wants to fence in a rectangular plot of land adjacent to the north wall of his barn. no fencing is needed along the bar

jeyben [28]

Hi,

Let assume a the west side,

b the length of the north wall (i suppose the answer is smaller then the barn.

Cost of the fence= a*6+a*12+b*12=3000 $.

So 3a+2b=500==>b=(500-3a)/2

Area =a*b=a(500-3a)/2= 250a-3/2a²

Derivate the area ==> 250-3a=0

==> a=250/3 and b=(500-3*250/3)/2=125

Width=125 (feet)

Length=250/3=83.333 (feet)

3 0

2 years ago

A half-century ago, the mean height of women in a particular country in their 20s was 64.7 inches. Assume that the heights of

Ainat [17]

Answer:

99.5% of all samples of 21 of today's women in their 20's have mean heights of at least 65.86 inches.

Step-by-step explanation:

We are given that a half-century ago, the mean height of women in a particular country in their 20's was 64.7 inches. Assume that the heights of today's women in their 20's are approximately normally distributed with a standard deviation of 2.07 inches.

Also, a samples of 21 of today's women in their 20's have been taken.

Let $\bar X$ = sample mean heights

The z-score probability distribution for sample mean is given by;

Z = $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean height of women = 64.7 inches

$\sigma$ = standard deviation = 2.07 inches

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, Probability that the sample of 21 of today's women in their 20's have mean heights of at least 65.86 inches is given by = P( $\bar X$ $\geq$ 65.86 inches)

P( $\bar X$ $\geq$ 65.86 inches) = P( $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ $\geq$ $\frac{65.86-64.7}{\frac{2.07}{\sqrt{21} } }$ ) = P(Z $\geq$ -2.57) = P(Z $\leq$ 2.57)