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kiruha [24]
2 years ago
15

PLSS HELP ITS DUE IN HALF AN HOURRR!!

Mathematics
1 answer:
natulia [17]2 years ago
6 0

Answer

A

Step-by-step explanation:

$\mathrm{Convert\:mixed\:numbers\:to\:improper\:fractions}:\quad6\frac{2}{3}=\frac{20}{3}$

=8\div \frac{20}{3}

=\frac{8}{1}\div \frac{20}{3}

=\frac{8}{1}\times \frac{3}{20}

=\frac{2}{1}\times \frac{3}{5}

=\frac{2\times \:3}{1\times \:5}

=\frac{6}{1\times \:5}

=\frac{6}{5}

Convert improper fractions to mixed numbers

=1\frac{1}{5}

Brainliest plz

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The women'schorus has 33members. How many quartetscan be formed? How many singers will be left over?
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Answer: 8 quartets can be formed.

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Step-by-step explanation:

A quartet is a group of 4 members.

Given: The women'schorus has 33 members.

The number of quartets can be formed = (Total members) divided by (4)

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By solving 33 ÷ 4  we get 8 as quotient and remainder as 1 such that

33 = 4 x 8 +1

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A horticulturist working for a large plant nursery is conducting experiments on the growth rate of a new shrub. Based on previou
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Answer:

t=\frac{1.8-2}{\frac{0.5}{\sqrt{48}}}=-2.771    

The degrees of freedom are given by:

df=n-1=48-1=47  

And the p value would be:

p_v =P(t_{(47)}    

Since the p value is lower than the significance level we have enough evidence to conclude that the true mean for this case for the growth rate is less than 2cm per week

Step-by-step explanation:

Information given

\bar X=1.8 represent the sample mean for the growth

s=0.5 represent the sample standard deviation    

n=48 sample size    

\mu_o =2 represent the value that we want to compare

\alpha=0.01 represent the significance level

t would represent the statistic    

p_v represent the p value

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean is less than 2cm per week, the system of hypothesis are :    

Null hypothesis:\mu \geq 2    

Alternative hypothesis:\mu < 2    

Since we don't know the population deviation the statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)    

Replacing the info given we got:

t=\frac{1.8-2}{\frac{0.5}{\sqrt{48}}}=-2.771    

The degrees of freedom are given by:

df=n-1=48-1=47  

And the p value would be:

p_v =P(t_{(47)}    

Since the p value is lower than the significance level we have enough evidence to conclude that the true mean for this case for the growth rate is less than 2cm per week

6 0
3 years ago
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