Help me out plz geometry

Which golf ball went higher, and how many feet? (Desmos!) - Just added the answer choices!

lora16 [44]

Answer:

Max height: 64 feet, and the socond one was higher.

Step-by-step explanation:

The max height is the y value of the vertex, because that’s when the graph peaks.

we can already very clearly see the vertex on the graph, so we don’t need to calculate it.

the max height of the second golf ball is 64 feet.

Now let’s look at the max height on the first golf ball.

we get the equation

h=-16t squared + 48t

to find the vertex of this, we can use the formula -b/2a

-48/-32 = 1.5

1.5 is the t value of this vertex.

to find the h value, we plug it in.

h = -16 (1.5) squared + 48(1.5)

h =2.25 times -16 + 72

h = -36 +72

h = 36

the first one is 36 max height, and the second is 64. The second one is bigger.

8 0

3 years ago

Mrs. Burnham has a 1-cup measure that has no other markings. Can she make 6 batches of modeling clay using on the 1-cup measure?

dexar [7]

Something along the lines of multiplying and adding

6 0

4 years ago

Read 2 more answers

Simplify square root of five times the quantity six minus four square root of three.

NemiM [27]

6√5 - 4√3-

i took the test but sorry if its wrong

5 0

3 years ago

HELP!!!!!! i need help with this thingy on my 5th grade hw but... idk and i'm a little to lazy to do this so here is the thingy

Makovka662 [10]

Assuming she did a checkered pattern if she got 10 red tiles she would need 20 tiles total if the kitchen was a square.
There seems to be some sort of missing information here or maybe a missing picture. Either way, I hope this helps.

4 0

3 years ago

The Institute of Management Accountants (IMA) conducted a survey of senior finance professionals to gauge members’ thoughts on g

DochEvi [55]

Answer:

(1) The probability that the sample percentage indicating global warming is having a significant impact on the environment will be between 64% and 69% is 0.3674.

(2) The two population percentages that will contain the sample percentage with probability 90% are 0.57 and 0.73.

(3) The two population percentages that will contain the sample percentage with probability 95% are 0.55 and 0.75.

Step-by-step explanation:

Let X = number of senior professionals who thought that global warming is having a significant impact on the environment.

The random variable X follows a Binomial distribution with parameters n = 100 and p = 0.65.

But the sample selected is too large and the probability of success is close to 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of p if the following conditions are satisfied:

np ≥ 10
n(1 - p) ≥ 10

Check the conditions as follows:

$np= 100\times 0.65=65>10\\n(1-p)=100\times (1-0.65)=35>10$

Thus, a Normal approximation to binomial can be applied.

So, $\hat p\sim N(p, \frac{p(1-p)}{n})=N(0.65, 0.002275)$ .

(1)

Compute the value of $P(0.64$ as follows:

$P(0.64$

$=P(-0.20$

Thus, the probability that the sample percentage indicating global warming is having a significant impact on the environment will be between 64% and 69% is 0.3674.

(2)

Let $p_{1}$ and $p_{2}$ be the two population percentages that will contain the sample percentage with probability 90%.

That is,

$P(p_{1}$

Then,

$P(p_{1}$

$P(\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}$

$P(-z$

The value of z for P (Z < z) = 0.95 is

z = 1.65.

Compute the value of $p_{1}$ and $p_{2}$ as follows:

$-z=\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}\\-1.65=\frac{p_{1}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{1}=0.65-(1.65\times 0.05)\\p_{1}=0.5675\\p_{1}\approx0.57$ $z=\frac{p_{2}-p}{\sqrt{\frac{p(1-p)}{n}}}\\1.65=\frac{p_{2}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{2}=0.65+(1.65\times 0.05)\\p_{1}=0.7325\\p_{1}\approx0.73$

Thus, the two population percentages that will contain the sample percentage with probability 90% are 0.57 and 0.73.

(3)

Let $p_{1}$ and $p_{2}$ be the two population percentages that will contain the sample percentage with probability 95%.

That is,

$P(p_{1}$

Then,

$P(p_{1}$

$P(\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}$

$P(-z$

The value of z for P (Z < z) = 0.975 is

z = 1.96.

Compute the value of $p_{1}$ and $p_{2}$ as follows:

$-z=\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}\\-1.96=\frac{p_{1}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{1}=0.65-(1.96\times 0.05)\\p_{1}=0.552\\p_{1}\approx0.55$ $z=\frac{p_{2}-p}{\sqrt{\frac{p(1-p)}{n}}}\\1.96=\frac{p_{2}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{2}=0.65+(1.96\times 0.05)\\p_{1}=0.748\\p_{1}\approx0.75$

Thus, the two population percentages that will contain the sample percentage with probability 95% are 0.55 and 0.75.

7 0

4 years ago