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irga5000 [103]
2 years ago
6

For the 1,000-meter running event, Abby's mean time is 4.5 minutes with a standard deviation of 0.75 minutes. For the 800-meter

event, Abby's mean time is 3.2 minutes with a standard deviation of 0.4 minutes. In the last track meet, Abby ran the 1,000-meter race in 4.1 minutes and the 800-meter race in 3 minutes. In which event did Abby have a better performance?
Mathematics
2 answers:
LiRa [457]2 years ago
8 0

Answer:

Abby had a better performance in the 800-meter because this z-score was closer to the mean than the z-score for the 1,000-meter.

Step-by-step explanation:

did it on egde

Lilit [14]2 years ago
3 0

Answer:

He performed better in the 800 m race

Step-by-step explanation:

Obtain the standardized score for each event :

Zscore = (score - mean) / standard deviation

1000 meter :

Mean = 4.5 ; Standard deviation = 0.75 ; x = 4.1

Zscore = (4.1 - 4.5) / 0.75

Zscore = - 0.4 / 0.75 = - 0.5333

800 meter :

Mean = 3.2 ; Standard deviation = 0.4 ; x = 3

Zscore = (3 - 3.2) / 0.4

Zscore = - 0.2 / 0.4 = - 0.5

-0.5 > - 0.533

Hence, he performed better in the 800 m race

You might be interested in
Performance task: A parade route must start And and at the intersections shown on the map. The city requires that the total dist
GaryK [48]

Answer:

Part A: The proposed route does not meet requirement because it is longer than the maximum required length of 3 miles

Part B: For the total distance is as close to 3 miles as possible, the start point of the parade should be at the point on Broadway with coordinates (9.941, 4.970)

Part C: The coordinates of the cameras stationed half way down each road are;

For central avenue; (4, 2)

For Broadway; (7.97, 2.49)

Step-by-step explanation:

Part A: The length of the given route can be found using the equation for the distance, l, between coordinate points as follows;

l = \sqrt{\left (y_{2}-y_{1}  \right )^{2}+\left (x_{2}-x_{1}  \right )^{2}}

Where for the Broadway potion of the parade route, we have;

(x₁, y₁) = (12, 3)

(x₂, y₂) = (6, 0)

l_1 = \sqrt{\left (0 -3\right )^{2}+\left (6-12 \right )^{2}} = 3 \cdot \sqrt{5}

For the Central Avenue potion of the parade route, we have;

(x₁, y₁) = (6, 0)

(x₂, y₂) = (2, 4)

l_2 = \sqrt{\left (4 -0\right )^{2}+\left (2-6 \right )^{2}} = 4 \cdot \sqrt{2}

Therefore, the total length of the parade route =-3·√5 + 4·√2 = 12.265 unit

The scale of the drawing is 1 unit = 0.25 miles

Therefore;

The actual length of the initial parade =0.25×12.265 unit = 3.09 miles

The proposed route does not meet requirement because it is longer than the maximum required length of 3 miles

Part B:

For an actual length of 3 miles, the length on the scale drawing should be given as follows;

1 unit = 0.25 miles

0.25 miles = 1 unit

1 mile =  1 unit/(0.25) = 4 units

3 miles = 3 × 4 units = 12 units

With the same end point and route, we have;

l_1 = \sqrt{\left (0 -y\right )^{2}+\left (6-x \right )^{2}} = 12 - 4 \cdot \sqrt{2}

y² + (6 - x)² = 176 - 96·√2

y² = 176 - 96·√2 - (6 - x)²............(1)

Also, the gradient of l₁ = (3 - 0)/(12 - 6) = 1/2

Which gives;

y/x = 1/2

y = x/2 ..............................(2)

Equating equation (1) to (2) gives;

176 - 96·√2 - (6 - x)² = (x/2)²

176 - 96·√2 - (6 - x)² - (x/2)²= 0

176 - 96·√2 - (1.25·x²- 12·x+36) = 0

Solving using a graphing calculator, gives;

(x - 9.941)(x + 0.341) = 0

Therefore;

x ≈ 9.941 or x = -0.341

Since l₁ is required to be 12 - 4·√2, we have and positive, we have;

x ≈ 9.941 and y = x/2 ≈ 9.941/2 = 4.97

Therefore, the start point of the parade should be the point (9.941, 4.970) on Broadway so that the total distance is as close to 3 miles as possible

Part C: The coordinates of the cameras stationed half way down each road are;

For central avenue;

Camera location = ((6 + 2)/2, (4 + 0)/2) = (4, 2)

For Broadway;

Camera location = ((6 + 9.941)/2, (0 + 4.970)/2) = (7.97, 2.49).

5 0
2 years ago
The owners of Prim's Pizza are concerned that many of their customers are starting to purchase pizza from The Pizza Palace becau
AleksandrR [38]

Answer:

(240-210) -1.984 \sqrt{\frac{8.6^2}{100}+\frac{5.7^2}{100}}=27.953

(240-210) +1.984 \sqrt{\frac{8.6^2}{100}+\frac{5.7^2}{100}}=32.047

And the confidence interval for the difference is between:

27.953 \leq \mu_1 -\mu_2 \leq 32.047

Step-by-step explanation:

We have the following info given:

\bar X_1 = 240 sample mean for medium Pizzas from Prim's

\bar X_2 = 210 sample mean for medium Pizzas from Pizza Place

s_1 =8.6 sample deviation for Prim's

s_2 =5.7 sample deviation for Pizza Palca

n_1 =n_2 = 100  sample size selected for each case

The confidence interval for the difference of means is given by:

(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}

And for the 95% confidence we need a significance level of \alpha=1-0.95=0.05 and \alpha/2 =0.025, the degrees of freedom are given by:

df= n_1 +n_2 -2= 100+100-2 =98

And the critical value would be

t_{\alpha/2}= 1.984

And replacing we got:

(240-210) -1.984 \sqrt{\frac{8.6^2}{100}+\frac{5.7^2}{100}}=27.953

(240-210) +1.984 \sqrt{\frac{8.6^2}{100}+\frac{5.7^2}{100}}=32.047

And the confidence interval for the difference is between:

27.953 \leq \mu_1 -\mu_2 \leq 32.047

8 0
2 years ago
According to the Rational Root Theorem, which could be a factor of the polynomial f(x) = 60x4 + 86x3 – 46x2 – 43x + 8?
kaheart [24]
<span> If P(x) is a polynomial with integer coefficients and if is a zero of P(x) ( P( ) = 0 ), then p is a factor of the constant term of P(x) and q is a factor of the leading coefficient of P(x) .

</span><span>A. x – 6
</span><span>60(6)^4 + 86(6)^3 – 46(6)^2 – 43(6) + 8 = 94430
</span><span>
B. 5x – 8
</span>60(8/5)^4 + 86(8/5)^3 – 46(8/5)^2 – 43(8/5) + 8 = 566.912<span>

C. 6x – 1
</span>60(1/6)^4 + 86(1/6)^3 – 46(1/6)^2 – 43(1/6) + 8 = 0 -------> ANSWER
<span>
D. 8x + 5
</span>60(-5/8)^4 + 86(-5/8)^3 – 46(-5/8)^2 – 43(-5/8) + 8 = 5.07
3 0
2 years ago
Read 2 more answers
Nathan is out rafting. He rafts 16 miles with the river current. At the end of 16 miles, he turns around and rafts the same dist
Novosadov [1.4K]
S = d/t

st = d

t = d/s

The time going is t1.
The time returning is t2.
The total time is 4 hours, so we have t1 + t2 = 4

The speed of the current is c.
The speed going is 9 + c.
The speed returning is 9 - c.

t1 = 16/(9 + c)

t2 = 16/(9 - c)

t1 + t2 = 16/(9 + c) + 16/(9 - c)

4 = 16/(9 - c) + 16/(9 + c)

1 = 4/(9 - c) + 4/(9 + c)

(9 + c)(9 - c) = 4(9 - c) + 4(9 + c)

81 - c^2 = 36 - 4c + 36 + 4c

81 - c^2 = 72

c^2 = 9

c^2 - 9 = 0

(c + 3)(c - 3) = 0

c + 3 = 0   or   c - 3 = 0

c = -3   or   c = 3

We discard the negative answer, and we get c = 3.

The speed of the current is 3 mph.
7 0
2 years ago
Read 2 more answers
Given a data set consisting of 33 unique whole number observations, its five-number summary is: [19,32,47,61,77] how many observ
Dafna11 [192]

Given a data set consisting of 33 unique whole number observations, its five-number summary is: [19,32,47,61,77] how many observations are strictly less than 32?

Solution: The five number summary denotes:

Minimum = 19

First Quartile = 32

Median = 47

Third Quartile = 61

Maximum = 77

Since there are an odd number of observations (33) in the data set, the First Quartile (32) must be at the (33+1)/4 = 8.5th position, meaning there are 7 numbers less than 32.

Therefore, there are 7 observations that are strictly less than 32

3 0
2 years ago
Read 2 more answers
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