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3 years ago

8

Ryan is taking a survey to find out how many people prefer soda A over soda B. He surveyed 100 people and forty-nine preferred s

oda A. Seven percent had no preference for soda A or soda B. What percentage of peoples surveyed preferred soda B?

1 answer:

marysya [2.9K]3 years ago

8 0

Answer:

44%

Step-by-step explanation:

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Over time, the value of the property at 397 West Lake Street increased by 275%. If the initial value of the property was P, whic

larisa86 [58]

Answer: Third option is correct.

Step-by-step explanation:

Since we have given that

Initial value of the property = P

Rate of growth = 275%

So, we need to find the property's current value:

We will use "compound interest ":

$Amount=P(1+\dfrac{r}{100})^t\\\\Amount=P(1+\dfrac{275}{100})^1\\\\A=P(1+2.75)\\\\A=3.75P$

Hence, Third option is correct.

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3 years ago

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2 years ago

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What is the sum of the interior angles of a regular polygon with 13 sides

Leviafan [203]

Answer:

Regular Polygons

Sides - Name - Interior Angles

13 - Triskaidecagon - 152.31°

14 - Tetrakaidecagon. - 154.29°

15 - Pendedecagon. - 156.00°

16 - Hexdecagon - 157.50°

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3 years ago

The Ray family went out to dinner The price of the meal was $38.95 The sales tax was 7% of the price of the meal. The tip was 18

Andrews [41]

Answer:

$49.18

Step-by-step explanation:

Dinner cost $38.95 before tax and tip are included.

Directions say to ADD 7% tip first so,

38.95 multiplied by 7%= 2.7265, Round up to $2.73

ADD $38.95 and $2.73 tax= $41.68

Now, multiply $41.68 by 18% to find the tip amount

$41.68 * 18%= 7.5024, Round to $7.50

ADD $41.68 and $7.50 to get answer: $49.18

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3 years ago

Quantitative noninvasive techniques are needed for routinely assessing symptoms of peripheral neuropathies, such as carpal tunne

Pavlova-9 [17]

Answer:

$t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507$

$p_v =P(t_{(15)}>1.507)=0.076$

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

Step-by-step explanation:

1) Data given and notation

$\bar X_{CTS}=2.35$ represent the mean for the sample CTS

$\bar X_{N}=1.83$ represent the mean for the sample Normal

$s_{CTS}=0.88$ represent the sample standard deviation for the sample of CTS

$s_{N}=0.54$ represent the sample standard deviation for the sample of Normal

$n_{CTS}=10$ sample size selected for the CTS

$n_{N}=7$ sample size selected for the Normal

$\alpha=0.01$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean for the group CTS is higher than the mean for the Normal, the system of hypothesis would be:

Null hypothesis: $\mu_{CTS} \leq \mu_{N}$

Alternative hypothesis: $\mu_{CTS} > \mu_{N}$

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{CTS}-\bar X_{N}}{\sqrt{\frac{s^2_{CTS}}{n_{CTS}}+\frac{s^2_{N}}{n_{N}}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n_{CTS}+n_{N}-2=10+7-2=15$

Since is a one side right tailed test the p value would be:

$p_v =P(t_{(15)}>1.507)=0.076$

We can use the following excel code to calculate the p value in Excel:"=1-T.DIST(1.507,15,TRUE)"

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

6 0

4 years ago