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Katyanochek1 [597]

3 years ago

12

What is the sum of 2.5 × 104 and 3.4 × 104?

2 answers:

Bond [772]3 years ago

7 0

260 + 353.6 = 613.6 that's your answer

steposvetlana [31]3 years ago

7 0

613.6 would be the answer. I got it by multiplying 2.5 and 104 together to get 260, then I multiplied 3.4 and 104 together getting 353.6. I then added 260 and 353.6 together to get 613.6.

2.5 * 104 = 260
3.4 * 104 = 353.6
353.6 + 260 = 613.6

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A tree casts a shadow of 24 feet at the same time as a 5-foot tall man casts a shadow of 4 feet Find the height of the tree

Mice21 [21]

Answer:

25 Feet

Step-by-step explanation:

If the man is 5 feet and he cast a shadow of 4 feet that means that there is a 1 foot difference. So, subtract the shadow hight by 1.

3 0

3 years ago

The manager of a supermarket would like to determine the amount of time that customers wait in a check-out line. He randomly sel

garri49 [273]

Complete question:

The manager of a supermarket would like to determine the amount of time that customers wait in a check-out line. He randomly selects 45 customers and records the amount of time from the moment they stand in the back of a line until the moment the cashier scans their first item. He calculates the mean and standard deviation of this sample to be barx = 4.2 minutes and s = 2.0 minutes. If appropriate, find a 90% confidence interval for the true mean time (in minutes) that customers at this supermarket wait in a check-out line

Answer:

(3.699, 4.701)

Step-by-step explanation:

Given:

Sample size, n = 45

Sample mean, x' = 4.2

Standard deviation $\sigma$ = 2.0

Required:

Find a 90% CI for true mean time

First find standard error using the formula:

$S.E = \frac{\sigma}{\sqrt{n}}$

$= \frac{2}{\sqrt{45}}$

$= \frac{2}{6.7082}$

$SE = 0.298$

Standard error = 0.298

Degrees of freedom, df = n - 1 = 45 - 1 = 44

To find t at 90% CI,df = 44:

Level of Significance α= 100% - 90% = 10% = 0.10

$t_\alpha_/_2_, _d_f = t_0_._0_5_, _d_f_=_4_4 = 1.6802$

Find margin of error using the formula:

M.E = S.E * t

M.E = 0.298 * 1.6802

M.E = 0.500938 ≈ 0.5009

Margin of error = 0.5009

Thus, 90% CI = sample mean ± Margin of error

Lower limit = 4.2 - 0.5009 = 3.699

Upper limit = 4.2 + 0.5009 = 4.7009 ≈ 4.701

Confidence Interval = (3.699, 4.701)

5 0

4 years ago

Andrei [34K]

Its the first one. Beacuse she gains speed then slows down and stops but then starts up again then slows down. Hope this helps!

7 0

4 years ago

Read 2 more answers

Surface area of this right prism

Sphinxa [80]

$\qquad\qquad\huge\underline{{\sf Answer}}$

$\textsf{Let's calculate the surface area of given right prism}$

$\textbf{Area of Triangles :}$

$\sf{\dfrac{1}{2}\cdot 16\cdot 12}$

$\sf{8×12}$

$\sf96 \: \: cm {}^{2}$

$\textsf{Since there are two triangles, }$

$\textsf{Area of Triangles = 96 × 2 = 192 cm²}$

$\textbf{Now, calculate the Areas of rectangles :}$

$\textsf{Area - 1 = 12 × 10 = 120 cm²}$

$\textsf{Area - 2 = 16 × 10 = 160 cm²}$

$\textsf{Area - 3 = 20 × 10 = 200 cm²}$

$\textsf{Now add them all :}$

$\textsf{120 + 160 + 200 + 192}$

$\textsf{672 cm²}$

7 0

3 years ago

Read 2 more answers

A pan is heated to 393°F, then removed from the heat and allowed to cool in a kitchen where the room temperature is a constant 6

deff fn [24]

Answer:

The approximate temperature of the pan after it has been away from the heat for 9 minutes is 275.59°F.

Step-by-step explanation:

The formula for D, the difference in temperature between the pan and the room after t minutes is:

$D = 324\cdot e^{-0.05t}$

Compute the approximate difference in temperature between the pan and the room after 9 minutes as follows:

$D = 324\cdot e^{-0.05t}$

$=324\times e^{-0.05\times 9}\\\\=206.59$

Then the approximate temperature of the pan after it has been away from the heat for 9 minutes is:

D = P - R

206.59 = P - 69

P = 206.59 + 69

P = 275.59°F

Thus, the approximate temperature of the pan after it has been away from the heat for 9 minutes is 275.59°F.

6 0

3 years ago