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andrey2020 [161]

3 years ago

7

Bradley plans to borrow the money needed to buy his PS5. If he borrows $600 at 7% simple interest for one year, how much interes

t will he pay?

1 answer:

marysya [2.9K]3 years ago

7 0

Answer:

42$

I think this is right hope this helps

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Vincent propses an idea to the student government to install a basket hoop along with a court marked with all the shooting lines

sladkih [1.3K]

I think your question is missed of key information, allow me to add in and hope it will fit the original one. Please have a look at the attached photo.

My answer:

Scale Drawing Lengths: . . / .

Actual Court Lengths . . .

Scale Factor: inch corresponds to ( ∙ ) inches, or inches, so the scale factor is .

Let = , represent the scale drawing lengths in inches, and represent the actual court lengths in inches. The -values must be converted from feet to inches.

To find actual length:

= =

() = inches, or feet

To find actual width:

= = ( )

= / ∙ /

= inches, or feet

The actual court measures feet by feet. Yes, the lot is big enough for the court Vincent planned. The court will take up the entire width of the lot.

5 0

3 years ago

Where it cuts off it says renting. Please help me.

pishuonlain [190]

Letter a I think is correct because if it's 2.5 per game plus 5 dollars per month then the equation is 2.5v+5 and for 4 games it is 2.5(4)+5 which equals 15 and 5(4) is 20 so letter a is corrext correct

7 0

3 years ago

Which function does the graph represent? What is the asymptote of the function?

Snowcat [4.5K]

Answer:

your new acc has not verify

plss try again!!

5 0

2 years ago

A laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighings. Scale readings in repeated we

weqwewe [10]

Answer:

99% confidence interval for the given specimen is [3.4125 , 3.4155].

Step-by-step explanation:

We are given that a laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighing. Scale readings in repeated weighing are Normally distributed with mean equal to the true weight of the specimen.

Three weighing of a specimen on this scale give 3.412, 3.416, and 3.414 g.

Firstly, the pivotal quantity for 99% confidence interval for the true mean specimen is given by;

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

where, $\bar X$ = sample mean weighing of specimen = $\frac{3.412+3.416+3.414}{3}$ = 3.414 g

$\sigma$ = population standard deviation = 0.001 g

n = sample of specimen = 3

$\mu$ = population mean

<em>Here for constructing 99% confidence interval we have used z statistics because we know about population standard deviation (sigma).</em>

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99 {As the critical value of z at 0.5% level

of significance are -2.5758 & 2.5758}

P(-2.5758 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.5758) = 0.99

P( $-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

P( $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

<u>99% confidence interval for</u> $\mu$ = [ $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $3.414-2.5758 \times {\frac{0.001}{\sqrt{3} } }$ , $3.414+2.5758 \times {\frac{0.001}{\sqrt{3} } }$ ]

= [3.4125 , 3.4155]

Therefore, 99% confidence interval for this specimen is [3.4125 , 3.4155].

6 0

3 years ago

Find the value of y in the given triangle

Andre45 [30]

Given:

In the given triangle,

With respect to y, Perpendicular = 10 cm and Base = 8 cm

To find the value of y.

Formula

By Trigonometric Ratio we get,

$tan\theta = \frac{Perpendicular}{Base}$

Now,

Putting the values of perpendicular and base we get,

$tan\theta = \frac{10}{8}$

or, $\theta = tan^{-1} (\frac{10}{8})$

or, $\theta = 51.34 ^\circ$

Rounding off to the nearest tenth, we have;

$\theta = 51.3 ^\circ$

Hence,

The value of y is 51.3°.

7 0

3 years ago