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3 years ago

HELPPPPP ASAPPPP

Mathematics

1 answer:

KATRIN_1 [288]3 years ago

5 0

Answer:

The maximum height of the volleyball is 12.25 feet.

Step-by-step explanation:

The height of the volleyball, h(t), is modeled by this equation, where t represents the time, in seconds, after that ball was set :

$h(t)=-16t^2+20t+6$ ....(1)

The volleyball reaches its maximum height after 0.625 seconds.

For maximum height,

Put $\dfrac{dh}{dt}=0$

Now put t = 0.625 in equation (1)

$h(t)=-16(0.625)^2+20(0.625)+6\\\\h(t)=12.25\ \text{feet}$

So, the maximum height of the volleyball is 12.25 feet.

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Suppose △BUG is an isosceles triangle. Find the measure of <B.

algol [13]

Answer:

B=44°

Step-by-step explanation:

x+10+3x+x=180

5x= 180-10

x=34

B= 34+10

B=44°

7 0

2 years ago

2. State whether the change is increasing or decreasing. Round your percent of change to the nearest integer percent. originally

marysya [2.9K]

Answer:

The change is increasing with a percentage of change of 64%

Step-by-step explanation:

Since we have the present grater than the original, then the change is increasing

Secondly, we want to calculate the percent of change

Mathematically , that will be ;

(present - original)/original * 100%

= (90-55)/55 * 100%

= 35/55 * 100% = 63.63 which is approximately 64%

7 0

3 years ago

Newborn males have weights with a mean of 3272.8 g and a standard deviation of 660.2 g. Newborn females have weights with a mean

Maksim231197 [3]

The correct option is: a female who weighs 1500 g

Explanation

Formula for finding the z-score is: $z= \frac{X-\mu}{\sigma}$

Newborn males have weights with a mean $(\mu)$ of 3272.8 g and a standard deviation $(\sigma)$ of 660.2 g.

So, the z-score for the newborn male who weighs 1500 g will be.......

$z(X=1500)=\frac{1500-3272.8}{660.2}=-2.685... \approx -2.69$

According to the normal distribution table, $P(z=-2.69)=0.0036 = 0.36\%$

Now, newborn females have weights with a mean $(\mu)$ of 3037.1 g and a standard deviation $(\sigma)$ of 706.3 g.

So, the z-score for the newborn female who weighs 1500 g will be.......

$z(X=1500)=\frac{1500-3037.1}{706.3}=-2.176... \approx -2.18$

According to the normal distribution table, $P(z=-2.18)=0.0146 = 1.46\%$

As we can see that the probability that a newborn female has weight of 1500 g is greater than newborn male, so a newborn female has the weight of 1500 g that is more extreme relative to the group from which he came.

5 0

3 years ago

Simplify: 20 - 2 × 8 + 42 A. 20 B. 160 C. 52 D. 26

Gnom [1K]

Answer: 46

Step-by-step explanation:

You can simplify this using PEMDAS:

Parentheses

Exponents

Multiplication/division

Addition/subtraction

Move through the steps in that order.

In this case, the first two steps are absent, so multiply first:

20 - 16 + 42

Then, add and subtract:

46.

5 0

3 years ago

Each contestant in the Hunger Games must be trained to compete. Suppose that the time it takes to train a contestant has mean 5

LiRa [457]

Answer:

11.51% probability that it will take less than 450 days to train all the contestants.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .