I need help, with these two answers..

Can you please help me. Go to my acc for more help for me :)

Angelina_Jolie [31]

The answer is 164. 5x4=20 and so on

8 0

3 years ago

The amounts a soft drink machine is designed to dispense for each drink are normally distributed, with a mean of 11.711.7 fluid

dexar [7]

Answer: 0.1499

Step-by-step explanation:

Given : The amounts a soft drink machine is designed to dispense for each drink are normally distributed, with mean $\mu=11.7\text{ fluid ounces}$

Standard deviation : $\sigma=0.2\text{ days}$

z-score : $z=\dfrac{X-\mu}{\sigma}$

For X = 11.5

$z=\dfrac{11.5-11.7}{0.2}\approx-1$

For X = 11.6

$z=\dfrac{11.6-11.7}{0.2}\approx-0.5$

Now, the probability that the drink is between 11.511.5 and 11.611.6 fluid ounces will be :-

$P(11.5$

Hence, the probability that the drink is between 11.511.5 and 11.611.6 fluid ounces =0.1499

5 0

3 years ago

Please help!! will mark brainliest

Tpy6a [65]

Answer:

The answer to your question my friend is Y = -1, 3 x

Step-by-step explanation:

I am making my best guest so that I can answer your question since no one will

Thx and have a blessed day

3 0

3 years ago

Each pound of food C contains 6 ounces of ingredient P and 3 ounces of ingredient Q. Each pound of food D contains 4.5 ounces of

GuDViN [60]

So hmm the food C = 6p + 3q and the food D = 4.5p + 1.5q

now, if we check how much is the ratios of C/D for each component, then, mixture M = 144p + 60q, must contain the same ratio for each "p" and "q" component

$\bf \textit{ratio of "p"}\implies \cfrac{6}{4.5}\implies \cfrac{4}{3}
\\\\\\
\textit{ratio of "q"}\implies \cfrac{3}{1.5}\implies \cfrac{2}{1}$

so, if we divide the 144p by 4+3, or 7 even pieces, 4 must belong to food C and 3 to food D, retaining the ratio of 4/3

and we do the same for 60q, we divide it in 2+1 or 3 even pieces, but that one is very clear, 2 must belong to food C and 1 to food D, 60/3 is clear ends up with a ratio of 40 and 20

now the "p" part... ends up as $\bf \cfrac{144}{7}\cdot 4\qquad and\qquad \cfrac{144}{7}\cdot 3\implies \cfrac{576}{7}\qquad and\qquad \cfrac{432}{7}$

that's what I see it, as the ratio of 4/3 and 2/1 being retained in the mixture M

7 0

3 years ago

At the end of the ride up a steep hill, Ken was at an elevation of 1600 meters above where he started. He figured out that he an

Lostsunrise [7]

Answer:

Step-by-step explanation:

first place)

in first place you must Know by definition that a joule is defined as the amount of work done by a constant force in a meter of length

and it is 1 J = 1 N ×m = 1 kg ×m/s² ×m= 1 Kg×m²/s²

now you can calculate the time done by ken and his bicycle

1000000 = 52kg×(1600)²/s² = 1000000= 133120000/s²⇒1000000s² = 133120000⇒s²= 133120000/1000000⇒ s² =133,12⇒s =√133,12 = 11,5 ≅12

12 s is the time that ken with his bicycle ride up to the mountain

second )

having the time you can calculate the work done by ken individually

52kg×(1600)²/(12)² =133120000/ (12)² = 924444,4 and this is the work done by ken

Now you must substract the work done by ken and his bicycle from the work done by ken individually in order to know the work of the bicycle

1000000 joules -924444,4 joules = 75555,6 joules

Having this amount, you can calculate the mass of the bicycle

and it is 1 J = 1 N ×m = 1 kg ×m/s² ×m= 1 Kg×m²/s²

75555,6 = kg *(1600) ²/(12)²⇒ 75555,6 =17777,7kg⇒ Kg = 75555,6/17777,7 = 4,25 kg

the mass of the ken's bicycle is 4,25 kg

3 0

3 years ago