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2 years ago

-) A floor plan shows that a tiny house has a length of 30 feet and a width of 20 feet.

Mathematics

1 answer:

Artyom0805 [142]2 years ago

8 0

If the first one is supposed to be 30 multiplied by 20 then that is the answer

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Rewrite the expression in the form x

Serga [27]

Answer:

See below (I hope this helps!)

Step-by-step explanation:

When dividing exponential terms of the same base, we can simply subtract the exponents, therefore:

$\frac{x^{6} }{x^{8} } = x^{(6-8)} = x^{-2}$

5 0

3 years ago

Help!

Vinvika [58]

A<b+c so if b=15 and c=40 then
a<15+40
a<45

so a cant possibly be d. 50ft

6 0

3 years ago

The box on the left is 8 inches wide and 8 inches tall what is the surface area of the box

TiliK225 [7]

Answer:

384 inches

Step-by-step explanation:

8 x 8 = 64

one face of the box is 64

6 faces of the box would be 384

6 0

3 years ago

Read 2 more answers

Solve for a.

Shalnov [3]

Answer:

the answer is -3

Step-by-step explanation:

1. add the -4 to 10 (-10 + 4)

2. -6= 2a

3. divide both sides by 2

4. -3= a

hope this helps

3 0

2 years ago

Read 2 more answers

There were 800 math instructors at a mathematics convention. Forty instructors were randomly selected and given an IQ test. The

Leya [2.2K]

Answer:

95% confidence interval for the mean of the 800 instructors is [126.80 , 133.20].

Step-by-step explanation:

We are given that there were 800 math instructors at a mathematics convention.

Forty instructors were randomly selected and given an IQ test. The scores produced a mean of 130 with a standard deviation of 10.

Firstly, the pivotal quantity for 95% 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 score = 130

s = sample standard deviation = 10

n = sample of instructors = 40

$\mu$ = population mean of 800 instructors

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

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

P(-2.0225 < $t_3_9$ < 2.0225) = 0.95 {As the critical value of t at 39 degree of

freedom are -2.0225 & 2.0225 with P = 2.5%}

P(-2.0225 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.0225) = 0.95

P( $-2.0225 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.0225 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.0225 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.0225 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-2.0225 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.0225 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $130-2.0225 \times {\frac{10}{\sqrt{40} } }$ , $130+2.0225 \times {\frac{10}{\sqrt{40} } }$ ]

= [126.80 , 133.20]

Therefore, 95% confidence interval for the mean of the 800 instructors is [126.80 , 133.20].

5 0

3 years ago

Read 2 more answers