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Colt1911 [192]
3 years ago
13

√15(√6+6) A. √21+6√15 B. 3√10+6√15 C. √90+6 D. √90+6√15

Mathematics
1 answer:
Varvara68 [4.7K]3 years ago
4 0

Answer:

Before A. 32.7247330577, A. 27.8204757722, B. 32.7247330577, C. 15.4868329805, and D. 32.7247330577

Step-by-step explanation:

Used a calculator. Not that Hard.

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272 sweets are divided between Ram and Rahim in the ratio of 7 : 9. How many sweets do they get?​
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Answer:

let the ratio of ram and hari be 7x and 9x respectively

7x +9x=272

or, 16x=272

or,x=272÷16

or,x=17

ratio of ram=7x=7 × 17=119

ratio pf rahim=9x=9×17=153

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Step-by-step explanation:

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Read 2 more answers
A marketing study was conducted to compare the mean age of male and femlae purchasers of a certain product. random and independe
notka56 [123]

Answer:

A. t ≥ 1.734

Step-by-step explanation:

Data given and notation  

\bar X_{m}=39.80 represent the mean for the sample male

\bar X_{f}=50.30 represent the mean for the sample female

s_{m}=10.040 represent the sample standard deviation for the males  

s_{f}=13.215 represent the sample standard deviation for the females  

n_{m}=10 sample size for the group male  

n_{f}=10 sample size for the group female  

t would represent the statistic (variable of interest)  

Concepts and formulas to use  

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

Null hypothesis:\mu_{f}-\mu_{m}\leq 0  

Alternative hypothesis:\mu_{f} - \mu_{m}> 0  

We don't have the population standard deviation, so for this case is better apply a t test to compare means, and the statistic is given by:  

t=\frac{(\bar X_{f}-\bar X_{m})-\Delta}{\sqrt{\frac{s^2_{f}}{n_{f}}+\frac{\sigma^2_{m}}{n_{m}}}} (1)

And the degrees of freedom are given by df=n_m +n_f -2=10+10-2=18  

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.

What is the test statistic?

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

t=\frac{(50.3-39.80)-0}{\sqrt{\frac{13.215^2}{10}+\frac{10.040^2}{16}}}}=2.001

Critical value

For this case since we have a right tailed test we need to look into the t distribution with 18 degrees of freedom a value that accumulates 0.05 of the area on the right, and on this case:

t_{crit}=1.734

And the rejection zone of the null hypothesis would be: A. t ≥ 1.734

For our case our calculated value is higher than the critical value so we have enough evidence to reject the null hypothesis at the 5% of significance.

8 0
4 years ago
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