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3 years ago

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One serving of pretzels contains 1.5 grams of fat. That is 3% of the daily amount recommended for a 2,000 calorie diet. How many

grams of fat are recommended? Show your work.

1 answer:

OleMash [197]3 years ago

8 0

Step-by-step explanation:

given that 1.5 gram of fat =3% of 2000 calorie diet.

Thus, 1.5 gram of fat contributes to 3/100*2000= 60 calorie of diet.

Generally 6% of the diet must be fat.(recommended)

Since, 1.5 gram = 3%

thus, 3 gram =6%.

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A ball dropped from the top of a building can be modeled by the function f(t)=-16^2+36, where t represents time in seconds

Viefleur [7K]

Answer:

The ball is dropped from a height of 36 feet

The bee and the ball will collide after approximately 1.3 seconds

The bee and the ball will collide at approximately 8 feet above the ground

The ball hits the ground after 1.5 seconds

Step-by-step explanation:

<u><em>The complete question is</em></u>

A ball dropped from the top of the building can be modeled by the function f(t)=-16t^2 + 36 , where t represents time in seconds after the ball was dropped. A bee's flight can be modeled by the function, g(t)=3t+4, where t represents time in seconds after the bee starts the flight.

The graph represents the situation.

select all that apply

1) The bee launches into flight from the ground.

2) The ball is dropped from a height of 36 feet.

3) The bee and the ball will collide after approximately 1.3 seconds.

4) The bee and the ball will collide after approximately 8 seconds.

5) The bee and the ball will collide at approximately 8 feet above the ground.

6) The bee and the ball will not collide.

7) The ball hits the ground after 1.5 seconds

see the attached figure to better understand the problem

<u><em>Verify all the options</em></u>

case 1) The bee launches into flight from the ground.

The statement is false

Because

we have

$g(t)=3t+4$

For t=0

$g(t)=3(0)+4=4\ ft$

That means ----> The bee launches into flight from a height 4 feet above the ground

case 2) The ball is dropped from a height of 36 feet

The statement is true

Because

For t=0

$f(t)=-16(0)^2+36=36\ ft$

case 3) The bee and the ball will collide after approximately 1.3 seconds

The statement is true

Because

Equate f(t) and g(t)

$-16t^2+36=3t+4$

$-16t^2-3t+32=0$

solve the quadratic equation by graphing

The solution is t=1.324 sec

see the attached figure N 2

case 4) The bee and the ball will collide after approximately 8 seconds

The statement is false

Because, the bee and the ball will collide after approximately 1.3 seconds (see case 3)

case 5) The bee and the ball will collide at approximately 8 feet above the ground

The statement is true

Because

For t=1.324 sec

substitute the value of t in f(t) or g(t)

$g(t)=3(1.324)+4=7.97\ ft$

case 6) The bee and the ball will not collide

The statement is false

see case 3) and case 5)

case 7) The ball hits the ground after 1.5 seconds

The statement is true

Because

we know that

The ball hit the ground when the value of f(t) is equal to zero

so

For f(t)=0

$-16t^2+36=0$

$t^2=\frac{36}{16}$

$t=\frac{6}{4}=1.5\ sec$

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Answer:

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From the given data,

$\sf \:\:\:\:\:\:\:\:\:\:x_{k}=35$
$\sf \:\:\:\:\:\:\:\:\:\:f_{k}=50$
$\sf \:\:\:\:\:\:\:\:\:\:f_{k-1}=34$
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$\sf \:\:\:\:\:\:\:\:\:\:h=5$

${\bf \:\: {By\:using\:the\: formula}} \\ \\$

$\:\dag\:{\small{\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}}} \\ \\$

$\sf \:\:\:\:\:\:\:\:\:= 35+ {\bigg(5 \times \dfrac{(50 - 34)}{ ( 2 \times 50 - 34 - 42)}\bigg)} \\ \\$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= 35 +{\bigg(5 \times \dfrac{16}{24}\bigg)} \\ \\$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= {\bigg(35+\dfrac{10}{3}\bigg)} \\ \\$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(35 + 3.33) =.38.33 \\ \\$

$\:\:\sf {Hence,}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ \large{\underline{\mathcal{\gray{ mode\:=\:38.33}}}} \\ \\$

${\large{\frak{\pmb{\underline{Additional\: information }}}}}$

MODE

Most precisely, mode is that value of the variable at which the concentration of the data is maximum.

MODAL CLASS

In a frequency distribution the class having maximum frequency is called the modal class.

${\bf{\underline{Formula\:for\: calculating\:mode:}}} \\$

${\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}} \\ \\$

Where,

$\sf \small\pink{ \bigstar} \: x_{k}= lower\:limit\:of\:the\:modal\:class\:interval.$

$\small \blue{ \bigstar}$ $\sf \: f_{k}=frequency\:of\:the\:modal\:class$

$\sf \small\orange{ \bigstar}\: f_{k-1}=frequency\:of\:the\:class\: preceding\:the\;modal\:class$

$\sf \small\green{ \bigstar}\: f_{k+1}=frequency\:of\:the\:class\: succeeding\:the\;modal\:class$

$\small \purple{ \bigstar}$ $\sf \: h= width \:of\:the\:class\:interval$

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