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dalvyx [7]
3 years ago
13

Giraffes are herbivores,or plant eaters. A giraffe can eat up to 75 pounds of leaves each day. Write and evaluate an expression

to find how many pounds of leaves 5 giraffes can eat in a week.​
Mathematics
1 answer:
Rudik [331]3 years ago
7 0

Answer: 2625

Let’s say you multiply 5 giraffes by 75 pounds = 375lbs 5 giraffes eat a day multiply the 5 giraffes and they answer should be 2625lbs

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Find the arc length of the semicircle?
sukhopar [10]

Answer:

5(pi) (exactly) or 15.7 (approximately)

Step-by-step explanation:

The length of arc of an entire circle is the circumference of the circle.

The length of arc of a semicircle circle is half of the circumference of the circle.

C = pi * d

For a semicircle:

arc length = 0.5 * pi * d

arc length = 0.5 * pi * 10

arc length = 5(pi)

arc length = 15.7

6 0
3 years ago
The Westwood Warriors basketball team wants to score more points. To get better at scoring points the team is trying to improve
Makovka662 [10]

Answer:

It is better for the warriors to use man-to-man defense.

Step-by-step explanation:

The complete question is: The Westwood Warriors basketball team wants to score more points. To get better at scoring points the  team is trying to improve its offensive strategies. Some opponents primarily use a zone defense, while  others primarily use a man-to-man defense. When the Warriors play against teams that use a zone defense they score an average of 67 points per game with a standard deviation of 8 points per game. When they play against teams that use a  man-to-man defense they score an average of 62 points per game with a standard deviation of 5 points per game.

Since the Warriors started using their improved offensive strategies they have played two  games with the following results.

Against the McNeil Mavericks

Maverick defense: zone

Warrior points: 77

Against the Round Rock Dragons

Dragon defense: man-to-man

Warrior points: 71

What is the Z-score of these values?

We are given that when the Warriors play against teams that use a zone defense they score an average of 67 points per game with a standard deviation of 8 points per game. When they play against teams that use a  man-to-man defense they score an average of 62 points per game with a standard deviation of 5 points per game.

We have to find the z-scores.

  • Finding the z-score for the zone defense;

Let X = <u><em>points score by warriors when they use zone defense</em></u>

The z-score probability distribution for the normal distribution is given by;

                            Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = mean score = 67 points

            \sigma = standard deviation = 8 points

It is stated that the Warriors scored 77 points when they used zone defense, so;

   z-score for 77 =  \frac{X-\mu}{\sigma}

                            =  \frac{77-67}{8}  = <u>1.25</u>

  • Finding the z-score for the zone defense;

Let X = <u><em>points score by warriors when they use man-to-man defense</em></u>

The z-score probability distribution for the normal distribution is given by;

                            Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = mean score = 62 points

            \sigma = standard deviation = 5 points

It is stated that the Warriors scored 71 points when they used man-to-man defense, so;

   z-score for 71 =  \frac{X-\mu}{\sigma}

                            =  \frac{71-62}{5}  = <u>1.8</u>

<u></u>

So, it is better for the warriors to use man-to-man defense.

3 0
3 years ago
A book claims that more hockey players are born in January through March than in October through December. The following data sh
astra-53 [7]

Answer:

\chi^2 = \frac{(67-47.5)^2}{47.5}+\frac{(56-47.5)^2}{47.5}+\frac{(30-47.5)^2}{47.5}+\frac{(37-47.5)^2}{47.5}=18.295

Now we can calculate the degrees of freedom for the statistic given by:

df=(categories-1)=4-1=3

And we can calculate the p value given by:

p_v = P(\chi^2_{3} >18.295)=0.00038

Since the p value is very low we have enough evidence to reject the null hypothesis and we can conclude that the players' birthdates are not uniformly distributed throughout the​ year

Step-by-step explanation:

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is no difference of birthdates distributed throughout the​ year

H1: There is a difference between birthdates distributed throughout the​ year

The level of significance assumed for this case is \alpha=0.05

The statistic to check the hypothesis is given by:

\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}

The table given represent the observed values, we just need to calculate the expected values with the following formula E_i = \frac{total}{4}

And replacing we got:

E_{1} =\frac{67+56+30+37}{4}=47.5

And now we can calculate the statistic:

\chi^2 = \frac{(67-47.5)^2}{47.5}+\frac{(56-47.5)^2}{47.5}+\frac{(30-47.5)^2}{47.5}+\frac{(37-47.5)^2}{47.5}=18.295

Now we can calculate the degrees of freedom for the statistic given by:

df=(categories-1)=4-1=3

And we can calculate the p value given by:

p_v = P(\chi^2_{3} >18.295)=0.00038

Since the p value is very low we have enough evidence to reject the null hypothesis and we can conclude that the players' birthdates are not uniformly distributed throughout the​ year

3 0
4 years ago
Hurry please also idk y brainly is making me type 20 characters
Mama L [17]

Answer:

Probably because they don't want a whole paragraph on some answers

8 0
4 years ago
Read 2 more answers
Tood knows that the ratio of boys and girls in his class is 3:5. Since 12 of the students are boys, he says there must be 36 stu
Vesna [10]

Answer:

=99

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
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