Ask question

Ask question

All categories

3 years ago

14

HELP! Vince weighs 160 pounds, and his friend Nadir weighs 140 pounds. Nadir calculated that his weight on another planet would

be about 56 lb. Approximately what would Vince weigh on the other planet?

2 answers:

kobusy [5.1K]3 years ago

6 0

Answer:

64

Step-by-step explanation:

140/56=2.5

160/2.5=64

Reika [66]3 years ago

4 0

160/140 = x/56
cross multiply
(140)(x) = (160)(56)
140x = 8960
x = 8960/140
x = 64 lbs <== Vince's weight on the other planet

You might be interested in

Question 3(Multiple Choice Worth 2 points)

Lilit [14]

The equivalent expression is seven-fifths squared times one-third cubed. Then the correct option is D.

<h3>What is an equivalent expression?</h3>

The equivalent is the expression that is in different forms but is equal to the same value.

Five-sevenths squared times one-third raised to the power of negative three all raised to the power of negative one.

Convert the sentence into numerical form.

$\rm \rightarrow \left [ \left (\dfrac{5}{7} \right )^2 \times \left ( \dfrac{1}{3}\right )^{-3} \right ] ^{-1}$

Simplify the expression, then we have

$\rm \rightarrow \left [ \left (\dfrac{7}{5} \right )^{2} \times \left ( \dfrac{1}{3}\right )^{3} \right ]$

The equivalent expression is seven-fifths squared times one-third cubed. Then the correct option is D.

More about the equivalent link is given below.

brainly.com/question/889935

#SPJ1

6 0

2 years ago

What is the result of 31.978 subtracted from 66.4?

jasenka [17]

<h2>Answer:</h2>

34.422

<h3>Explanation:</h3>

A splendiferous teen willing to help,

who listens to "Fight Song,"

stay salty...

5 0

3 years ago

Read 2 more answers

D=h+9 The variable H represents the number of hours worked, and the variable d represents the amount of dough prepared how many

Marta_Voda [28]

Answer:

12 hours

Step-by-step explanation:

If we assume that the units of D are kg, then we can put the given number into the equation and solve.

21 = h +9

12 = h . . . . . subtract 9

Eli worked 12 hours to prepare 21 kg of dough.

8 0

4 years ago

I don’t get this :(

muminat

Ok so for the first one it’s super simple I’m gonna do it as less complicated so it can help you :)

So first we need to convert the mixed number to an improper fraction

1 4/5 = 9/5

Now we need to reduce the numbers greatest common divisor which is 3

3/5 x 6

Now multiply the numbers

Your answer is 18/5 I hoped this help

Now for the second one

this one is a little more complicated and I don’t know how to make it sound easier so I’m gonna tell you the answer which is 580/21

The last one is the same steps we did for the first one which is 27/20

I hoped this helped you it is a little difficult but you’ll get the hang of it :)

5 0

3 years ago

Let the number of chocolate chips in a certain type of cookie have a Poisson distribution. We want the probability that a cookie

ludmilkaskok [199]

Answer:

$\lambda \geq 6.63835$

Step-by-step explanation:

The Poisson Distribution is "a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event".

Let X the random variable that represent the number of chocolate chips in a certain type of cookie. We know that $X \sim Poisson(\lambda)$

The probability mass function for the random variable is given by:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

And f(x)=0 for other case.

For this distribution the expected value is the same parameter $\lambda$

$E(X)=\mu =\lambda$

On this case we are interested on the probability of having at least two chocolate chips, and using the complement rule we have this:

$P(X\geq 2)=1-P(X$

Using the pmf we can find the individual probabilities like this:

$P(X=0)=\frac{e^{-\lambda} \lambda^0}{0!}=e^{-\lambda}$

$P(X=1)=\frac{e^{-\lambda} \lambda^1}{1!}=\lambda e^{-\lambda}$

And replacing we have this:

$P(X\geq 2)=1-[P(X=0)+P(X=1)]=1-[e^{-\lambda} +\lambda e^{-\lambda}[]$

$P(X\geq 2)=1-e^{-\lambda}(1+\lambda)$

And we want this probability that at least of 99%, so we can set upt the following inequality:

$P(X\geq 2)=1-e^{-\lambda}(1+\lambda)\geq 0.99$

And now we can solve for $\lambda$

$0.01 \geq e^{-\lambda}(1+\lambda)$

Applying natural log on both sides we have:

$ln(0.01) \geq ln(e^{-\lambda}+ln(1+\lambda)$

$ln(0.01) \geq -\lambda+ln(1+\lambda)$

$\lambda-ln(1+\lambda)+ln(0.01) \geq 0$

Thats a no linear equation but if we use a numerical method like the Newthon raphson Method or the Jacobi method we find a good point of estimate for the solution.

Using the Newthon Raphson method, we apply this formula:

$x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)}$

Where :

$f(x_n)=\lambda -ln(1+\lambda)+ln(0.01)$

$f'(x_n)=1-\frac{1}{1+\lambda}$

Iterating as shown on the figure attached we find a final solution given by:

$\lambda \geq 6.63835$

4 0

3 years ago