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gladu [14]
3 years ago
14

Problems; e.g. 200 bottles of equal capacity hold 350 litres of water.

Mathematics
2 answers:
spayn [35]3 years ago
7 0

Answer:

each bottle will hold 1.75 litre water.

zheka24 [161]3 years ago
6 0

Answer:

1.75liters per bottle

Step-by-step explanation:

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Overline MD cong overline LS additional information is necessary to show that triangle MTD cong triangle LGS by SSS?
aivan3 [116]

Answer:

TD \cong GS

Step-by-step explanation:

See comment for complete question

Given:

TM \cong GL

MD \cong LS

Required

The information that shows \triangle MTD \cong \triangle LGS by SSS

By SSS implies that, the three sides of both triangles are congruent

Already, we have:

TM \cong GL

MD \cong LS

The third side of \triangle MTD is TD

The third side of \triangle LGS is GS

So, for both to be congruent by SSS, the third sides must be congruent

i.e.

TD \cong GS

4 0
2 years ago
Audrey is buying a nee car for $32,998. She plans to make a down payment of $4,200. If she's to make a monthly payments of $525
Airida [17]
4,200+525(5×12)
4,200+525(60)=
4,200+31,500=35,700
6 0
3 years ago
Disregarding wind resistance the distance a body falls from rest varies directly as th squares of the time it falls if a skydive
Firdavs [7]

Answer:

The skydiver will fall 1600 ft in 10 seconds.

Step-by-step explanation:

<u>Distance a body falls from rest varies directly as the squares of the time it falls </u>

Let distance = s    

      time = t

Therefore , s = k \times t^{2}

Given, the body falls 64 ft in 2 seconds.

Substituting in above equation,

64 = k \times (2)^{2}

64 = k \times (4)

 k  = \frac{64}{4}

 k = 16 ft/s^{2}

We have to find how much he will fall in 10 seconds.

s = k \times t^{2}

s = 16 \times 10^{2}

s = 16 \times 100

s = 1600 ft

3 0
3 years ago
One bar of candy A and two bars of candy B have 774 calories. Two bars of candy A and one bar of candy B contains 786 calories.
vampirchik [111]
◆ Define the variables:

Let the calorie content of Candy A = a
and the calorie content of Candy B = b

◆ Form the equations:

One bar of candy A and two bars of candy B have 774 calories. Thus:

a + 2b = 774

Two bars of candy A and one bar of candy B contains 786 calories

2a + b = 786

◆ Solve the equations:

From first equation,
a + 2b = 774
=> a = 774 - 2b

Put a in second equation
2×(774-2b) + b = 786
=> 2×774 - 2×2b + b = 786
=> 1548 - 4b + b = 786
=> -3b = 786 - 1548
=> -3b = -762
=> b = -762/(-3) = 254 calorie

◆ Find caloric content:

Caloric content of candy B = 254 calorie

Caloric content of candy A = a = 774 - 2b = 774 - 2×254 = 774 - 508 = 266 calorie
5 0
3 years ago
2.A production process manufactures items with weights that are normally distributed with mean 10 pounds and standard deviation
Vesna [10]

Answer:

Step-by-step explanation:

Given that:

population mean = 10

standard deviation = 0.1

sample mean = 9.8 < x > 10.2

The z score can be computed as:

z = \dfrac{\bar x - \mu}{\sigma}

if x > 10.2

z = \dfrac{10.2- 10}{0.1}

z = \dfrac{0.2}{0.1}

z = 2

If x < 9.8

z = \dfrac{9.8- 10}{0.1}

z = \dfrac{-0.2}{0.1}

z = -2

The p-value = P (z ≤ 2) + P (z ≥ 2)

The p-value = P (z ≤ 2) + ( 1 -  P (z ≥ 2)

p-value = 0.022750 +(1 -   0.97725)

p-value = 0.022750 +  0.022750

p-value = 0.0455

Therefore; the probability of defectives  = 4.55%

the probability of acceptable = 1 - the probability of defectives

the probability of acceptable = 1 - 0.0455

the probability of acceptable = 0.9545

the probability of acceptable = 95.45%

4.55% are defective or 95.45% is acceptable.

sampling distribution of proportions:

sample size n=1000

p = 0.0455

The z - score for this distribution at most 5% of the items is;

z = \dfrac{0.05 - 0.0455}{\sqrt{\dfrac{0.0455\times 0.9545}{1000}}}

z = \dfrac{0.0045}{\sqrt{\dfrac{0.04342975}{1000}}}

z = \dfrac{0.0045}{\sqrt{4.342975 \times 10^{-5}}}

z = 0.6828

The p-value = P(z ≤ 0.6828)

From the z tables

p-value = 0.7526

Thus, the probability that at most 5% of the items in a given batch will be defective = 0.7526

The z - score for this distribution for at least 85% of the items is;

z = \dfrac{0.85 - 0.9545}{\sqrt{\dfrac{0.0455\times 0.9545}{1000}}}

z = \dfrac{-0.1045}{\sqrt{\dfrac{0.04342975}{1000}}}

z = −15.86

p-value = P(z ≥  -15.86)

p-value = 1 - P(z <  -15.86)

p-value = 1 - 0

p-value = 1

Thus, the probability that at least 85% of these items in a given batch will be acceptable = 1

6 0
3 years ago
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