Work out $3740 as a percentage of $4400

Dan is helping to build a stage for a concert.

Fudgin [204]

1,512 meters squared

4 0

2 years ago

A rectangular tank measuring 35 cm by 28 cm by 16 cm is 2/5 filled with

Alina [70]

Answer:

Amount of water left in tank = 440 cm³

Step-by-step explanation:

Given:

Sides of tank respectively = 35 cm , 28 cm , 16 cm

Water level in tank = 2/5

Side of a cubic tank = 18 cm

Find:

Amount of water left in tank

Computation:

Volume level in rectangular tank = [2/5][35 x 28 x 16]

Volume level in rectangular tank = [2/5][15,680]

Volume level in rectangular tank = [31,360/5]

Volume level in rectangular tank = 6,272 cm³

Volume of cube = side³

Volume of cube = 18³

Volume of cube = 5,832 cm³

Amount of water left in tank = Volume level in rectangular tank - Volume of cube

Amount of water left in tank = 6,272 - 5,832

Amount of water left in tank = 440 cm³

7 0

2 years ago

Debra and Ian shared in $1,000,000 estate. If Ian received $125,000 and debra the rest, what fraction of the estate did Debra re

choli [55]

Debra received 7/8
if you divide 1,000,000 by 125,000 you get 8. which means Ian got 1/8 of the money. leaving 7/8 for Debra

5 0

2 years ago

Read 2 more answers

On this graph, 4:00 p.m. occurs at 16 hours after midnight, and 6:00 p.m. occurs at 18 hours after midnight. Which statements ar

KatRina [158]

Answer:

TRUE OPTIONS ARE:

"The temperature increased until 4:00 p.m. "

"The temperature decreased after 6:00 p.m. "

"The temperature increased more quickly between 12:00 p.m. and 4:00 p.m. than before 12:00 p.m."

Step-by-step explanation:

graph attached and complete question below:

Which statements are true about the temperatures Luis recorded on the graph? Check all that apply.

The temperature increased until 4:00 p.m.
The temperature was not recorded between 4:00 p.m. and 6:00 p.m.
The temperature decreased after 6:00 p.m.
The temperature increased and then decreased before holding constant.
The temperature increased more quickly between 12:00 p.m. and 4:00 p.m. than before 12:00 p.m.

Until 1600 hours, the graph increases, so we can say temperature increased until 4.00 pm (FIRST OPTION TRUE).

From 1600 to 1800 hours (4 - 6pm), the temperature stayed same (horizontal line). This doesn't mean the temperature wasn't recorded. (2nd OPTION FALSE).

After 1800 hours (6pm), the line goes downward, so temperature decreased after 6 pm. (3rd OPTION TRUE).

If you look at the temperature graph, we can see temperature increased, then increased more, then constant, then decreased. Thus the 4th option isnt true. (4th OPTION FALSE).

Before 12, the increase isn't as sharp as after 12. After 12 temperature increase has more slope, so this increase is more. (5th OPTION TRUE).

7 0

2 years ago

Read 2 more answers

A large manufacturing plant has analyzed the amount of time required to produce an electrical part and determined that the times

WARRIOR [948]

Answer:

We conclude that the new procedure will not decrease the population mean amount of time required to produce the part.

Step-by-step explanation:

We are given that a large manufacturing plant has analyzed the amount of time required to produce an electrical part and determined that the times follow a normal distribution with mean time μ = 45 hours.

A random sample of 25 parts will be selected and the average amount of time required to produce them will be determined. The sample mean amount of time is = 43.118 hours with the sample standard deviation s = 5.5 hours

Let $\mu$ = population mean amount of time required to produce an electrical part using new procedure

SO, Null Hypothesis, $H_0$ : $\mu \geq$ 45 hours {means that the new procedure will remain same or increase the population mean amount of time required to produce the part}

Alternate Hypothesis, $H_a$ : $\mu$ < 45 hours {means that the new procedure will decrease the population mean amount of time required to produce the part}

The test statistics that will be used here is One-sample t test statistics because we don't know about the population standard deviation;

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

where, $\mu$ = sample mean amount of time = 43.118 hours

s = sample standard deviation = 5.5 hours

n = sample of parts = 25

So, test statistics = $\frac{43.118-45}{{\frac{5.5}{\sqrt{25} } } }$ ~ $t_2_4$

= -1.711

Now at 0.025 significance level, the t table gives critical value of -2.064 at 24 degree of freedom for left-tailed test. Since our test statistics is more than the critical value of t so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the new procedure will remain same or increase the population mean amount of time required to produce the part.

4 0

2 years ago