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kvv77 [185]
3 years ago
5

5. A race course is 99 meters long. There are trees

Mathematics
2 answers:
Ivanshal [37]3 years ago
7 0

Answer:

61

Step-by-step explanation:

99 minus 38

Svet_ta [14]3 years ago
5 0

Answer:

61 meters

Step-by-step explanation:

To find this answer, you need to subtract the length with trees from the total length of the course. With this information, you get an equation that looks something like x + 38 = 99, with x being the length of the course without trees.

Subtract 38 from both sides:

x + 38 = 99

  - 38   - 38

x = 99 - 38

Simplify:

x = 61

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To test Upper H 0​: muequals50 versus Upper H 1​: muless than50​, a random sample of size nequals23 is obtained from a populatio
natta225 [31]

Answer:

Step-by-step explanation:

Hello!

1)

<em>To test H0: u= 50 versus H1= u < 50, a random sample size of n = 23 is obtained from a population that is known to be normally distributed. Complete parts A through D. </em>

<em> A) If  ¯ x = 47.9  and s=11.9, compute the test statistic .</em>

For thistest the corresponsing statistis is a one sample t-test

t= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } }~~t_{n-1}

t_{H_0}= \frac{47.9-50}{\frac{11.9}{\sqrt{23} } } = -0.846= -0.85

B) If the researcher decides to test this hypothesis at the a=0.1 level of significance, determine the critical value(s).

This test is one-tailed to the left, meaning that you'll reject the null hypothesis to small values of the statistic. The ejection region is defined by one critical value:

t_{n-1;\alpha }= t_{22;0.1}= -1.321

Check the second attachment. The first row shows α= Level of significance; the First column shows ν= sample size.

The t-table shows the values of the statistic for the right tail. P(tₙ≥α)

But keep in mind that this distribution is centered in zero, meaning that the right and left tails are numerically equal, only the sign changes. Since in this example the rejection region is one-tailed to the left, the critical value is negative.

C) What does the distribution graph appear like?

Attachment.

D) Will the researcher reject the null hypothesis?

As said, the rejection region is one-tailed to the right, so the decision rule is:

If t_{H_0} ≤ -1.321, reject the null hypothesis.

If t_{H_0} > -1.321, do not reject the null hypothesis.

t_{H_0}= -0.85, the decision is to not reject the null hypothesis.

2)

To test H0​: μ=100 versus H1​:≠​100, a simple random sample size of nequals=24 is obtained from a population that is known to be normally distributed. Answer parts​ (a)-(d).

a) If x =104.2 and s=9.6, compute the test statistic.

For this example you have to use a one sample t-test too. The formula of the statistic is the same:

t_{H_0}= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } } = \frac{104.2-100}{\frac{9.6}{\sqrt{24} } = } = 2.143

b) If the researcher decides to test this hypothesis at the α=0.01 level of​ significance, determine the critical values.

This hypothesis pair leads to a two-tailed rejection region, meaning, you'll reject the null hypothesis at either small or big values of the statistic. Then the rejection region is divided into two and determined by two critical values (the left one will be negative and the right one will be positive but the module of both values will be equal).

t_{n-1;\alpha/2 }= t_{23; 0.005}= -2.807

t_{n-1;1-\alpha /2}= t_{23;0.995}= 2.807

c) Draw a​ t-distribution that depicts the critical​ region(s). Which of the following graphs shows the critical​ region(s) in the​t-distribution?

Attachment.

​(d) Will the researcher reject the null​ hypothesis?

The decision rule for the two-tailed hypotheses pair is:

If t_{H_0} ≤ -2.807 or if t_{H_0} ≥ 2.807, reject the null hypothesis.

If -2.807 < t_{H_0} < 2.807, do not reject the null hypothesis.

t_{H_0}= 2.143 is greater than the right critical value, the decision is to reject the null hypothesis.

Correct option:

B. The researcher will reject the null hypothesis since the test statistic is not between the critical values.

3)

Full text in attachment. The sample size is different by 2 but it should serve as a good example.

H₀: μ = 20

H₁: μ < 20

a) n= 18, X[bar]= 18.3, S= 4, Compute statistic.

t_{H_0}= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } }= \frac{18.3-20}{\frac{4}{\sqrt{18} } } = -1.80

b) The rejection region in this example is one-tailed to the left, meaning that you'll reject the null hypothesis to small values of t.

Out of the three graphics, the correct one is A.

c)

To resolve this you have to look for the values in the t-table that are the closest to the calculated t_{H_0}

Symbolically:

t_{n-1;\alpha_1 } \leq t_{H_0}\leq t_{n-1;\alpha _2}

t_{H_0}= -1.80

t_{17; 0.025 }= -2.110

t_{17;0.05}= -1.740

Roughly defined you can say that the p-value is the probability of obtaining the value of t_{H_0}, symbolically: P(t₁₇≤-1.80)

Under the distribution the calculated statistic is between the values of -2.110 and -1.740, then the p-value will be between their cumulated probabilities:

A. 0.025 < p-value < 0.05

d. The researcher decides to test the hypothesis using a significance level of α: 0.05

Using the p-value approach the decision rule is the following:

If p-value ≤ α, reject the null hypothesis.

If p-value > α, do not reject the null hypothesis.

We already established in item c) that the p-value is less than 0.05, so the decision is to reject the null hypothesis.

Correct option:

B. The researcher will reject the null hypothesis since the p-value is less than α.

I hope this helps!

6 0
3 years ago
James Wants to triple his vegetarian meatball recipe that uses 1 lb. 10 oz. of tofu how much tofu will he need in total
AVprozaik [17]
Answer: The tripled tofu is 4 lb. 14 oz. Explanation: So you first want to turn your mass into the same unit, so change the pounds of tofu into ounces. There are 16 ounces in a pound, which means the total mass of the original tofu could be written as (16 oz.) + 10 oz. = 26 oz. tofu. Multiply this by 3, and you have tripled the tofu. 3 * 26 = 78. To simplify it, divide 78 oz by 16 to find pounds, and leave the remainder in ounces. 78/16 = 4 lb. 14 oz tofu.
7 0
3 years ago
Two airlines offer special group rates to your school’s Spanish Club for a trip to Mexico City. Mexican Air airline offers a rou
Finger [1]

Answer:

a. The Mexican Airline offers a better deal for 9 students to travel.  

b. 10 students

Step-by-step explanation:

a. The Mexican Air airline offers a round trip airfare of $250 per person.

So, for 9 students the fare will be total $(250 × 9) = $2250

Now, the Fiesta airline offers a round trip airfare of $150 per person if the club agrees to pay a one-time group rate processing fee of $1000.

So, for 9 students the total fare will be = $[1000 + (150 × 9)] =$2350.

Therefore, the Mexican Airline offers a better deal for 9 students to travel.  (Answer)

b. Let for x students both the airline offers the same total costs.

Then we can write  

250x = 1000 + 150x {From the conditions given}

⇒ 100x = 1000

⇒ x = 10

So, if the number of students is 10, then only both the airlines will take same fare. (Answer)

8 0
3 years ago
HELP PLEASE!!! I am doing my math and I need someone to help me with this problem. I am very confused and my family is gone on a
yanalaym [24]

Answer:

1) 250 meters= 0.25 km

2) 12 meters=  0.012km

3) 1 meter= 0.001 km

The constant of proportionality is 1000 meters in 1km (1000 meters)

I think there’s a mistake on the second part because 1 meter does not equal to 1000km. but I did part 1 for you!

4 0
3 years ago
A football team loses 3 yards in 3 consecutive plays. What’s the total yardage gained?
Anni [7]
-3×3=-9 so gain is 0
4 0
3 years ago
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