All categories

3 years ago

David’s father bought 5 smoothies. He paid $20. Write and solve an equation to find the cost of one smoothie.

Mathematics

2 answers:

Naya [18.7K]3 years ago

7 0

Answer:

Step-by-step explanation:

Because he bought 5 smoothies and they were each $20, you have to do 20÷5. Then you get $4 as your answer.

scoray [572]3 years ago

5 0

David’s father bought 5 smoothies. He paid $20. Write and solve an equation to find the cost of one smoothie . 5 / 20 = $4

You might be interested in

Need some help, please. What do I plot? I attempted to figure the questions out many times, but I've come to no clear conclusion

olganol [36]

Answer:

vertex is (4,-4) and another point is (6,0) or you could use (2,0) or many other options :)

Step-by-step explanation:

The cool thing about this question your quadratic is in factored form so your x-intercepts are easy to figure out, they are 2 and 6.

So you can plot (6,0) and (2,0).

The vertex will lie half between x=2 and x=6... so it lays at (6+2)/2=4

We just have to find the y-coordinate for when x=4.

Plug in 4 gives you (4-2)(4-6)=(2)(-2)=-4.

So the vertex is at (4,-4).

5 0

3 years ago

What is (13x + 9k) + (17x + 6k), simplified?

Mazyrski [523]

The answer is 15x+30k

7 0

3 years ago

Look at the table. Make a conjecture about the sum of the first 20 positive even numbers.

Ludmilka [50]

It will be 20*21 based on the pattern in the chart

3 0

2 years ago

Read 2 more answers

Who's the main charecter in Naruto answer gets Brainliest <br><br>i know what i spelt wrong

Ne4ueva [31]

Answer:

naruto uzumaki is the main character in the naruto

3 0

3 years ago

Read 2 more answers

Question 6 2 pts

astraxan [27]

Answer:

785 acres

Step-by-step explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Let X represent the random variable amount of corn yield

Assuming the X follows a normal distribution

$X \sim N(\mu, \sigma)$

And we know that the average $\bar X$ is dsitributed:

$\bar X \sim N(\mu=189.3, \sigma_{\bar X}=23.5)$

And we are interested on this probability:

$P(\bar X>180)$

For this case we can use the z score formula given by:

$z=\frac{\bar X -\mu}{\sigma_{\bar X}}$

If we apply this we got:

$P(\bar X>180)=P(Z>\frac{180-189.3}{23.5})=P(Z>-0.396)=1-P(Z$

And since we have a proportion estimated and we hava a total of 1200 acres the expected to yield more than 180 bushels of corn per acre would be:

$r=1200*0.654=784.8$

And if we round up this amount we got $\approx 785$ and that would be the best option for this case.

5 0

2 years ago