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4 years ago

Solve x^2 = 64, where x is a real number.

2 answers:

7 0

$\text{Solve for x:}\\\\x^2 = 64\\\\\text{Square root both sides to cancel the exponent}\\\\\sqrt(x)=\pm\sqrt64\\\\x=8\,\,or\,\,x=-8$

abruzzese [7]4 years ago

4 0

Answer: x=8

Step-by-step explanation:

8×8=64

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3 years ago

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Students in a representative sample of 69 second-year students selected from a large university in England participated in a stu

Serhud [2]

Answer:

95% confidence interval estimate of μ, the mean procrastination scale for second-year students at this terval college is [39.34 , 42.66].

Step-by-step explanation:

We are given that for the 69 second-year students in the study at the university, the sample mean procrastination score was 41.00 and the sample standard deviation was 6.89.

Firstly, the pivotal quantity for 95% confidence interval for the true mean is given by;

P.Q. = $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean procrastination score = 41

s = sample standard deviation = 6.89

n = sample of students = 69

$\mu$ = population mean estimate

Here for constructing 95% confidence interval we have used One-sample t test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the true mean, $\mu$ is ;

P(-1.9973 < $t_6_8$ < 1.9973) = 0.95 {As the critical value of t at 68 degree

of freedom are -1.9973 & 1.9973 with P = 2.5%}

P(-1.9973 < $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ < 1.9973) = 0.95

P( $-1.9973 \times{\frac{s}{\sqrt{n} } }$ < ${\bar X -\mu}$ < $1.9973 \times{\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.9973 \times{\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.9973 \times{\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ =[ $\bar X-1.9973 \times{\frac{s}{\sqrt{n} } }$ , $\bar X+1.9973 \times{\frac{s}{\sqrt{n} } }$ ]

= [ $41-1.9973 \times{\frac{6.89}{\sqrt{69} } }$ , $41+1.9973 \times{\frac{6.89}{\sqrt{69} } }$ ]

= [39.34 , 42.66]

Therefore, 95% confidence interval estimate of μ, the mean procrastination scale for second-year students at this terval college is [39.34 , 42.66].

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4 years ago

What is the equation of the circle shown in the graph? 6 9 12 36

Naddika [18.5K]

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3 years ago

A rectangular container that has a length of 30 cm, a width of 20 cm, and a height of 24 cm is filled with water to a depth of 1

BaLLatris [955]

Answer:

$5400cm^{3}$ more water could be added to the container before the container overflows

Step-by-step explanation:

The volume of the rectangular container is got by multiplying its given dimensions: Length X breadth X height

This will be 30 X 20 X 24 = $14400 cm^{3}$

To find the volume of water inside the rectangular container, we will use the new height for our computation of volumes, keeping the length and breadth of the cylinder the same.

This will be $30 \times 20 \times 15 =9000 cm^{3}$

The volume of water needed to be added until the container overflows will be got by subtracting the volume of water present in the container from the total volume of the container

$14400 - 9000 =5400cm^{3}$

Note 1 $cm^{3}= 1 ml$

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3 years ago