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N76 [4]
3 years ago
7

Lowest denominator 1/15 9/5 3/8

Mathematics
2 answers:
yulyashka [42]3 years ago
7 0
Iwould say the answer is 120
Stels [109]3 years ago
3 0
<span>LCD(<span>115</span>, <span>95</span>, <span>38</span>) = LCM(15, 5, 8) = 23×3×5 = 120</span>
<span><span>115</span> = <span>8120</span></span><span><span>95</span> = <span>216120</span></span><span><span>38</span> = <span>45120
</span></span>
You might be interested in
A university administrator was interested in determining if there was a difference in the distance students travel to get from c
eduard

Answer:

1) Fail to reject the Null hypothesis

2) We do not have sufficient evidence to support the claim that the mean distance students traveled to school from their current residence was different for males and females.

Step-by-step explanation:

A university administrator wants to test if there is a difference between the distance men and women travel to class from their current residence. So, the hypothesis would be:

H_{o}: \mu_{F}-\mu_{M}=0\\\\ H_{a}: \mu_{F}-\mu_{M}\neq 0

The results of his tests are:

t-value = -1.05

p-value = 0.305

Degrees of freedom = df = 21

Based on this data we need to draw a conclusion about test. The significance level is not given, but the normally used levels of significance are 0.001, 0.005, 0.01 and 0.05

The rule of the thumb is:

  • If p-value is equal to or less than the significance level, then we reject the null hypothesis
  • If p-value is greater than the significance level, we fail to reject the null hypothesis.

No matter which significance level is used from the above mentioned significance levels, p-value will always be larger than it. Therefore, we fail to reject the null hypothesis.

Conclusion:

We do not have sufficient evidence to support the claim that the mean distance students traveled to school from their current residence was different for males and females.

7 0
3 years ago
In a certain computer, the probability of a memory failure is 0.01, while the probability of a hard disk failure is 0.02. If the
ratelena [41]

Answer:

We need to remember that we have independent events when a given event is not affected by previous events, and we can verify if two events are independnet with the following equation:

P(A \cap B) = P(A) *P(B)

For this case we have that:

P(A) *P(B) = 0.01*0.02= 0.0002

And we see that 0.0002 \neq P(A \cap B)

So then we can conclude that the two events given are not independent and have a relationship or dependence.

Step-by-step explanation:

For this case we can define the following events:

A= In a certain computer a memory failure

B= In a certain computer a hard disk failure

We have the probability for the two events given on this case:

P(A) = 0.01 , P(B) = 0.02

We also know the probability that the memory and the hard drive fail simultaneously given by:

P(A \cap B) = 0.0014

And we want to check if the two events are independent.

We need to remember that we have independent events when a given event is not affected by previous events, and we can verify if two events are independnet with the following equation:

P(A \cap B) = P(A) *P(B)

For this case we have that:

P(A) *P(B) = 0.01*0.02= 0.0002

And we see that 0.0002 \neq P(A \cap B)

So then we can conclude that the two events given are not independent and have a relationship or dependence.

8 0
2 years ago
Find the measures of
ki77a [65]

3y+15=4y-3

Subtract 3y from both sides

15=y-3

Add 3 to both sides

Y=18

4(18)-3

72-3

69 both angles are 69

8 0
3 years ago
What is 4/5 of 45 equle
Goryan [66]
\frac{4}{5} \ of \ 45= \frac{4}{5} \times 45=4 \times \frac{45}{5}=4 \times 9=36

4/5 of 45 equals 36.
5 0
3 years ago
Read 2 more answers
The mean throwing distance of a football for a Marco, a high school freshman quarterback, is 40 yards, with a standard deviation
Phoenix [80]

Answer:

a) One-tailed t-test about a mean.

b) The null hypothesis would be that there is no significant evidence to conclude that the adjusted grip leads to more distance on the thrown ball.

While the alternative hypothesis is that there is significant evidence to conclude that the adjusted grip leads to more distance on the thrown ball.

Mathematically,

The null hypothesis is represented as

H₀: μ ≤ 40 yards

The alternative hypothesis is given as

Hₐ: μ > 40 yards

c) p-value = 0.00621

d) The graph is presented in the attached image. The p-value obtained is less than the significance level at which the test was performed, hence, we accept the alternative hypothesis and say that there is significant evidence to conclude that the adjusted grip leads to more distance on the thrown ball.

Step-by-step explanation:

a) This test checks if the adjusted grip leads to more distance than before on the ball throws.

So, it is a one tailed test (testing only in one direction), a t-test about a mean.

b) For hypothesis testing, the first thing to define is the null and alternative hypothesis.

The null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.

While, the alternative hypothesis usually confirms the the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test.

Since this question aims to test if adjusting the grip leads to more distance on the ball thrown.

The null hypothesis would be that there is no significant evidence to conclude that the adjusted grip leads to more distance on the thrown ball. That is, The adjusted grip leads to a distance lesser than or equal to the previous distance on the thrown ball.

While the alternative hypothesis is that there is significant evidence to conclude that the adjusted grip leads to more distance on the thrown ball.

Mathematically,

The null hypothesis is represented as

H₀: μ ≤ 40 yards

The alternative hypothesis is given as

Hₐ: μ > 40 yards

c) To do this test, we will use the z-distribution because there is information on the population standard deviation.

So, we compute the test statistic

z = (x - μ₀)/σ

x = 45 yards

μ₀ = 40 yards

σ = standard deviation = 2 yards

z = (45 - 40)/2 = 2.50

The p-value for a one-tailed test for z-test statistic of 2.50 is 0.00621

d) The sketch of the graph is presented in the attached image.

The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 0.05

p-value = 0.00621

0.00621 < 0.05

Hence,

p-value < significance level

This means that we reject the null hypothesis & accept the alternative hypothesis and say that there is enough evidence to conclude that the adjusted grip leads to more distance on the thrown ball.

Hope this Helps!!!

5 0
3 years ago
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