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ki77a [65]
3 years ago
11

Jamal ran 8 kilometers sprinted 500 meters and climbed bleachers for a distance of 1.5 kilometers during track practice. How man

y meters in total did he travel during practice?
Mathematics
1 answer:
Sindrei [870]3 years ago
7 0

Answer:

Total distance traveled = 10000 meters

Step-by-step explanation:

1000 meters = 1 kilometers

Hence 8 kilometers = 8000 meters and 1.5 kilometers = 1500 meters

Total distance traveled = 10000 meters

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8 1/2 is to 17 as 17 11 1/2 is to 23
Stella [2.4K]

Hello from MrBillDoesMath!

Answer:   Yes,    (8 1/2) /`17 = ( 17 11/12) /23

** I think you meant  11 1/2 is to 23  NOT   17 11 1/2 is to 23"


Discussion:

First,  8 1/2 =   17/2 and

(17/2 ) / 17 = 1/2

Second  11 1/2 = 23/2 and

(23/2)/ 23 = 1/2

IN other words, the ratiors are the same:


8 1/2                 11 1/2

-------      =         -------

17                      23



Thank you,

MrB

5 0
3 years ago
At the fast food restaurant, an order of fries costs $1.14 and a drink costs $1.22. How
natima [27]

Answer:

$8.22, 1.14f + 1.22d

Step-by-step explanation:

1.14(4) + 1.22(3) = 4.56 + 3.66 = 8.22

1.14f + 1.12d

8 0
2 years ago
If Pooh Bear spots a bee hive in a tree at an angle of elevation of 48 degrees, then what is the angle of depression of the Quee
faltersainse [42]

Answer:

∅ = 90 - 48 = 42°

Step-by-step explanation:

4 0
3 years ago
Ackerman and Goldsmith (2011) report that students who study from a screen (phone, tablet, or computer) tended to have lower qui
enot [183]

Answer:

The scores are significantly lower than those from the general population.

Step-by-step explanation:

Hello!

To make the test we need to first identify the hypothesis we want to test. In this case, the hypothesis statement is

<em>"Studying from a screen lowers the scores on the final exam"</em>

Should this happen, it would mean that the average scores on the final exam will be lowered too. If this statement is not true, the average scores on the final exam should not change whether the students use virtual or printed materials to study.

On the other hand, we will take the previously known information as population reference, so for this example, the population mean is 577 and the standard deviation 58

With this in mind, we can state the null and alternative hypothesis:

H₀: μ = 577

H₁: μ < 577

The text doesn't specify a significance level, so I'll use the most common one. α=0.05

For this text, since we have a large sample (n=516), the variable has a normal distribution and its parameters known, we'll use a Z-test.

Z= (x(bar)-μ)/(σ/√n) ≈ N(0;1)

Critical region.

The rejection region is one-tailed, this is depicted in the hypothesis since it says the scores "lower" when virtual materials are used to study. So we will reject the null hypothesis if the calculated Z-value is less than the critical value.

Our critical value bein a Z_{\alpha } = Z_{\0.05} = -1.64

So we will reject the null hypothesis if the Z_{obs} is ≤-1.64 or support the null hypothesis if the Z_{obs}is >-1.64

Next we calculate the Z-value

Z_{obs}= (x(bar)-μ)/(σ/√n) = (572.5-577)/(58/√516)= -4.5/2.55 = -1.76

since Z_{obs}= -1.76 ≤ -1.64 we will reject the null hypothesis.

In other words, we can assume that the average scores on the final exam decrease when the students use virtual materials to study.

I hope you have a SUPER day!

4 0
3 years ago
Find the area of the polygon​
kobusy [5.1K]

Answer:    192

Step-by-step explanation:

You multiply 6 x 16 x 2

7 0
3 years ago
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