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3 years ago

Help me please (look the picture)

Mathematics

1 answer:

umka2103 [35]3 years ago

4 0

You add 1 to the numerator each time

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The following table shows the amount of lemon flavor and water needed in four punch recipes. Punch Recipes Lemon flavor Water Re

Sauron [17]

Answer:

recipe 2

Step-by-step explanation:

4 0

2 years ago

Please, I need help fast!

Vesna [10]

Answer:

Step-by-step explanation:

The answer to your question is -40

8 0

3 years ago

Read 2 more answers

A marine biologist will collect data regarding the length and weight of each of 50 trout caught and released at five different l

Vesnalui [34]

I think that the best visual representation that the marine biologist would use is utilising a bar graph. A bar graph is composed of bars residing in a coordinate plane. In addition to that, the visual representation would most likely be used since the average weight and length of each trout coming from 5 different locations is compared.

3 0

3 years ago

If (-2/15)+(-13/5)=(-13/5)+a/b then a/b is equal to

stellarik [79]

The choose C. –2/15

(-2/15) +(-13/5) =(-13/5) +a/b

a/b = (-2/15)+(-13/5) - (-13/5)

a/b = (-2/15) - ( 39/15) +(39/15)

a/b = –2/15

I hope I helped you ^_^

6 0

3 years ago

Read 2 more answers

A recent study reported that high school students spend an average of 94 minutes per day texting. Jenna claims that the average

larisa [96]

Answer:

We have sufficient evidence to support the claim that the average for the students at Jenna's large high school is greater than 94 minutes.

Step-by-step explanation:

Jenna claims that the average time of texting at her larger high school is greater than 94 minutes per day.

From here we can see that we have to perform a hypothesis test about a sample mean. The null and alternate hypothesis will be:

Null Hypothesis: $\mu \leq 94$

Alternate Hypothesis: $\mu > 94$

Jenna collected data from a sample of 32 students. So, sample size will be:

Sample Size = n = 32

Sample Mean = x = 96.5

Sample Standard Deviation = s = 6.3

We have to perform a hypothesis test, to test Jenna's claim. Since, the value of Population Standard Deviation is unknown and the value of Sample Standard Deviation is known, we will use One Sample t-test in this case.

The formula to calculate the test statistic is:

$t=\frac{x-\mu}{\frac{s}{\sqrt{n}}}$

Using the values, we get:

$t=\frac{96.5-94}{\frac{6.3}{\sqrt{32} } }=2.245$

The degrees of freedom will be:

df = n - 1 = 32 - 1 = 31

We have to convert the t-score 2.245 with 31 degrees of freedom to its equivalent p-value. From t-table this value comes out to be:

p-value = 0.0160

The significance level is:

$\alpha =0.05$

Since, the p-value is lesser than the level of significance, we reject the Null Hypothesis.

Conclusion:

We have sufficient evidence to support the claim that the average for the students at Jenna's large high school is greater than 94 minutes.

3 0

4 years ago