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solong [7]
3 years ago
5

The recommended weight of a soccer ball is 430 grams. The actual weight is allowed to vary by up to 20 grams.

Mathematics
1 answer:
Stella [2.4K]3 years ago
6 0
The recommended weight of a soccer ball is 430 grams. The actual weight is allowed to vary by up to 20 grams.
a. Write and solve an absolute value equation to find the minimum and maximum acceptable soccer ball weights.
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the greater roadrunner bird can run 14 miles per hour that's 7 times faster than an ostrich can walk how fast does an ostrich wa
Verizon [17]

Answer: An ostrich can walk with a speed of 2 miles per hour.


Step-by-step explanation:

Speed of roadrunner bird=14 miles per hour

Let x be the speed of ostrich egg.

Given that : A the greater roadrunner bird can run 7 times faster than an ostrich can walk.

⇒ Speed of roadrunner bird=7x

⇒7x= 14 miles per hour.

\\\Rightarrow\ x=\frac{14\text{ miles per hour}}{7}\\\\\Rightarrow\ x=2\text{ miles per hour}

Therefore, speed of ostrich egg=2 miles per hour.

5 0
3 years ago
In triangle NQL, point S is the centroid, NS = (x + 10) feet, and SR = (x + 3) feet.
olya-2409 [2.1K]
Where R is the median between Q and L:

From my understanding of a triangle's centroid, it divides an angle bisector into parts of 2/3 and 1/3. In the given problem, these divisions are NS and SR. Therefore, twice SR would be equal to NS. From here, we can get the value of X, to solve for SR.

NS = 2SR
(x + 10) = 2(x + 3)
x + 10 = 2x + 6
x = 4

Therefore, SR = (x + 3) = 7
4 0
3 years ago
Read 2 more answers
HELP PLZZ I WILL REWARD IF ANSWERS RIGHT
pentagon [3]
I know you have 1 and 3 right but I'm sorry about not knowing the rest.
8 0
3 years ago
A random sample of 100 automobile owners in thestate of Virginia shows that an automobile is driven onaverage 23,500 kilometers
Anastaziya [24]

Answer:

a) 22497.7 < μ< 24502.3

b)  With 99% confidence the possible error will not exceed 1002.3

Step-by-step explanation:

Given that:

Mean (μ) = 23500 kilometers per year

Standard deviation (σ) = 3900 kilometers

Confidence level (c) = 99% = 0.99

number of samples (n) = 100

a) α = 1 - c = 1 - 0.99 = 0.01

\frac{\alpha }{2} =\frac{0.01}{2}=0.005\\ z_{\frac{\alpha }{2}}=z_{0.005}=2.57

Using normal distribution table, z_{0.005 is the z value of 1 - 0.005 = 0.995 of the area to the right which is 2.57.

The margin of error (e) is given as:

e= z_{0.005}\frac{\sigma}{\sqrt{n} }  = 2.57*\frac{3900}{\sqrt{100} } =1002.3

The 99% confidence interval = (μ - e, μ + e) = (23500 - 1002.3, 23500 + 1002.3) =  (22497.7, 24502.3)

Confidence interval = 22497.7 < μ< 24502.3

b) With 99% confidence the possible error will not exceed 1002.3

5 0
3 years ago
A survey found that​ women's heights are normally distributed with mean 63.2 in. and standard deviation 2.4 in. The survey also
Effectus [21]

Answer: 99.51%

Step-by-step explanation:

Given : A survey found that​ women's heights are normally distributed.

Population mean : \mu =63.2  \text{ inches}

Standard deviation: \sigma= 2.4\text{ inches}

Minimum height = 4ft. 9 in.=4\times12+9\text{ in.}=57\text{ in.}

Maximum height = 6ft. 2 in.=6\times12+2\text{ in.}=74\text{ in.}

Let x be the random variable that represent the women's height.

z-score : z=\dfrac{x-\mu}{\sigma}

For x=57, we have

z=\dfrac{57-63.2}{2.4}\approx-2.58

For x=74, we have

z=\dfrac{74-63.2}{2.4}\approx4.5

Now, by using the standard normal distribution table, we have

The probability of women meeting the height requirement :-

P(-2.58

Hence, the percentage of women meeting the height requirement = 99.51%

8 0
3 years ago
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