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3 years ago

Helpppppppppppppppppppppppppppp

Mathematics

2 answers:

elena-s [515]3 years ago

8 0

$Given : \sqrt{64}$

$\implies \sqrt{64} = \sqrt{8\times8} = \sqrt{8^2} = (8^2)^\frac{1}{2} = 8^\frac{2}{2} = 8$

$Hence : \sqrt{64} = 8$

nevsk [136]3 years ago

4 0

1. Write the number in exponential form with a base of 8.

2. Reduce the index of the radical and exponent with 2.

3. Your answer is 8

I hope this helps☺

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Taryn conducted a science experiment on saturation. She added sugar to sugar water solution at different intervals. The graph sh

dusya [7]

Equation of a line:
$y=mx+c$

m = gradient: The difference between two y points and two x points.
$m= \frac{\Delta y}{\Delta x}$

c = y-intercept: Where the line crosses the y-axis (x=0)

You have:
$y=\textunderscore\textunderscore \ x + \textunderscore$
so you are missing the m and the c.

To calculate m find two y coordinates -you have (12, <u>7</u>) and (0, <u>1</u>)- and subtract them. Then divide this by the subtracted values of the x coordinates -you have (<u>12</u>, 7) and (<u>0</u>, 1)- This gives:

$m= \frac{7-1}{12-0}$
$m= \frac{6}{12}$
$m=0.5$

To calculate the c, you just see where the line crosses the y-axis. Because you have the point (0, 1), you know that when x=0, y=1. Because x=0 is on the y-axis, you can tell that the line passes through y=1. This makes your c = 1:

$c=1$

When you plug these values into the equation you get your answer:

$y=0.5x+1$

6 0

3 years ago

-1 1/5 + 3/4 I’m bad with fractions and don’t know how to do these it’s due tomorrow

Dovator [93]

-1 1/5 + 3/4 = -0.45

1/5 is 0.2 in decimal form, and 3/4 is 0.75:

-1.2 + 0.75 = -0.45

Hope this helps!

8 0

3 years ago

Read 2 more answers

One day at football practice, Darrell, the kicker, punted the ball so that its height in feet above the ground was given by the

nika2105 [10]

When the football is 81 feet high, it will be √ 3 / 8 seconds

<h3>Function</h3>

Function relates input to output. Therefore,

h(t) = - 16t² + 75

Therefore,

t = number of seconds

h(t) = - 16t² + 75

h(t) = 81 ft

Therefore,

81 = - 16t² + 75

- 16t² = 81 - 75

- 16t² = 6

divide both sides by -16

t² = 6 / -16

t² = - 3 / 8

square root both sides

t = √ 3 / 8 seconds

learn more on function here: brainly.com/question/4418459

6 0

2 years ago

How much acetic acid is in a 5 - liter container of acetic acid and water that is marked 80% acetic acid ? How much is water ?

Goryan [66]

I think the question ur asking is what’s the volume of water?? If so then the volume of water is 1L

6 0

1 year ago

A survey was conducted that asked 1003 people how many books they had read in the past year. Results indicated that x overbar eq

Sergio [31]

Answer:

The 99% confidence interval would be given (11.448;14.152).

Step-by-step explanation:

1) Important concepts and notation

A confidence interval for a mean "gives us a range of plausible values for the population mean. If a confidence interval does not include a particular value, we can say that it is not likely that the particular value is the true population mean"

$s=16.6$ represent the sample deviation

$\bar X=12.8$ represent the sample mean

n =1003 is the sample size selected

Confidence =99% or 0.99

$\alpha=1-0.99=0.01$ represent the significance level.

2) Solution to the problem

The confidence interval for the mean would be given by this formula

$\bar X \pm z_{\alpha/2} \frac{s}{\sqrt{n}}$

We can use a z quantile instead of t since the sample size is large enough.

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

And replacing into the confidence interval formula we got:

$12.8 - 2.58 \frac{16.6}{\sqrt{1003}}=11.448$

$12.8 + 2.58 \frac{16.6}{\sqrt{1003}} =14.152$

And the 99% confidence interval would be given (11.448;14.152).

We are confident that about 11 to 14 are the number of books that the people had read on the last year on average, at 1% of significance.

3 0

3 years ago