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3 years ago

The area of a square is 98 m2. How long is the diagonal?

1 answer:

N76 [4]3 years ago

6 0

A= sides2
98=sides2
√98 = √(〖sides〗^2 )
Side= 9.8994
Now the Pythagorean theorem
a2 + b2 = c2
9.89942 + 9.89942 = c2
97.99812036+ 97.99812036 = c2
c2 = 195.99624072
c= 13.99
Rounded to the nearest whole number is 14
A. 14 m

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What is the solution to the system of equation below Y equals -1/3 X +6 and X equals -6

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Answer:

y = 8

Step-by-step explanation:

y= $\frac{1}{3} x$ + 6

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2 years ago

Matthew scored a total of 168 points in basketball this season. He scored 147 of those in the regular season, and and the rest w

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12.5%

Step-by-step explanation:

Step1. Determine the points Matthew had during playoff by subtracting points made during regular season from total points in season.

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Step 2. Get the percentage, by dividing playoff points over total points in season

21 ÷ 168 = 0.125

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2 years ago

in a class of 30 students the highest score in physics test was 98 and the lowest was 35 what was the range?

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3 years ago

Boxes of raisins are labeled as containing 22 ounces. Following are the weights, in the ounces, of a sample of 12 boxes. It is r

ZanzabumX [31]

Answer:

A 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

Step-by-step explanation:

We are given the weights, in the ounces, of a sample of 12 boxes below;

Weights (X): 21.88, 21.76, 22.14, 21.63, 21.81, 22.12, 21.97, 21.57, 21.75, 21.96, 22.20, 21.80.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

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

where, $\bar X$ = sample mean weight = $\frac{\sum X}{n}$ = 21.88 ounces

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 0.201 ounces

n = sample of boxes = 12

$\mu$ = population mean weight

Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.796 < $t_1_1$ < 1.796) = 0.90 {As the critical value of t at 11 degrees of

freedom are -1.796 & 1.796 with P = 5%}

P(-1.796 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.796) = 0.90

P( $-1.796 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.796 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.796 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.796 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.796 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.796 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $21.88-1.796 \times {\frac{0.201}{\sqrt{12} } }$ , $21.88+1.796 \times {\frac{0.201}{\sqrt{12} } }$ ]

= [21.78, 21.98]

Therefore, a 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

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3 years ago

Find the expression to (x-1)(x-9)

disa [49]

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3 years ago

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