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il63 [147K]
3 years ago
7

Lettuce is packaged four bags to a box. If there are 3.5 boxes of lettuce in the cooler.How many bags of lettuce would there be?

Mathematics
1 answer:
Alenkasestr [34]3 years ago
8 0

Number of bags of lettuce in a box = 4

Number of boxes in the cooler = 3.5 boxes

So, number of bags of lettuce are = 4\times3.5 =14

Hence, there are 14 bags of lettuce in 3.5 boxes.

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Rebecca went swimming yesterday. After a while she had covered one fifth of her intended distance. After swimming six more lengt
alukav5142 [94]

Answer:

Option E.

Step-by-step explanation:

Let the intended length of Rebecca's swimming is = x units

and we assume the length of the pool = l units

Now it is given in the question that " She covers one fifth of her intended distance "

That means distance covered = \frac{x}{5}

" After swimming six more lengths of the pool she had covered one quarter of her intended distance"

So \frac{x}{5}+6(l)=\frac{x}{4}

6l=\frac{x}{4}-\frac{x}{5}

6l=\frac{x}{20}

x = 20×(6l)

x = 120l

Therefore, Rebecca has to complete 120 lengths of the pool.

Option E is the answer.

4 0
3 years ago
The graph of the equation -5x + By=18 is a line that passes through (-6,-2)
g100num [7]

Answer:

6

Step-by-step explanation:

B(Y)=5x+18

input the values

-2B=5(-6)+18

-2B=-30+18

-2B=-12

B=-12/-2

B=6

7 0
2 years ago
PLEASE HELP!! BRAINLIEST= 50 POINTS!!! This morning, some dragons, horses, and chickens were playing in my backyard. I counted 1
aev [14]
2 dragons
10 horses
5 chickens
7 0
3 years ago
Read 2 more answers
A random sample of 500 registered voters in Phoenix is asked if they favor the use of oxygenated fuels year-round to reduce air
Stells [14]

Answer:

a) 0.0853

b) 0.0000

Step-by-step explanation:

Parameters given stated that;

H₀ : <em>p = </em>0.6

H₁ : <em>p  = </em>0.6, this explains the acceptance region as;

p° ≤ \frac{315}{500}=0.63 and the region region as p°>0.63 (where p° is known as the sample proportion)

a).

the probability of type I error if exactly 60% is calculated as :

∝ = P (Reject H₀ | H₀ is true)

   = P (p°>0.63 | p=0.6)

where p° is represented as <em>pI</em><em> </em>in the subsequent calculated steps below

   

    = P  [\frac{p°-p}{\sqrt{\frac{p(1-p)}{n}}} >\frac{0.63-p}{\sqrt{\frac{p(1-p)}{n}}} |p=0.6]

    = P  [\frac{p°-0.6}{\sqrt{\frac{0.6(1-0.6)}{500}}} >\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500}}} ]

    = P   [Z>\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500} } } ]

    = P   [Z > 1.37]

    = 1 - P   [Z ≤ 1.37]

    = 1 - Ф (1.37)

    = 1 - 0.914657 ( from Cumulative Standard Normal Distribution Table)

    ≅ 0.0853

b)

The probability of Type II error β is stated as:

β = P (Accept H₀ | H₁ is true)

  = P [p° ≤ 0.63 | p = 0.75]

where p° is represented as <em>pI</em><em> </em>in the subsequent calculated steps below

  = P [\frac{p°-p} \sqrt{\frac{p(1-p)}{n} } }\leq \frac{0.63-p}{\sqrt{\frac{p(1-p)}{n} } } | p=0.75]

  = P [\frac{p°-0.6} \sqrt{\frac{0.75(1-0.75)}{500} } }\leq \frac{0.63-0.75}{\sqrt{\frac{0.75(1-0.75)}{500} } } ]

  = P[Z\leq\frac{0.63-0.75}{\sqrt{\frac{0.75(1-0.75)}{500} } } ]

  = P [Z ≤ -6.20]

  = Ф (-6.20)

  ≅ 0.0000 (from Cumulative Standard Normal Distribution Table).

6 0
3 years ago
Find the mean.<br> 20, 5, 45, 90, 60, 45, 30, 10, 20, 45, 15,25
aliina [53]

\huge\textsf{Hey there!}

\bullet \large\textsf{ Finding the mean you have to add up all of your numbers \& divide it}\\\large\textsf{by the total numbers in your data plot}

\mathsf{\dfrac{20 + 5 + 45 +  90 + 60 + 45 + 30 + 10 + 20+ 45 + 15 + 25}{12}}

\mathsf{20 + 5 + 45 +  90 + 60 + 45 + 30 + 10 + 20+ 45 + 15 + 25}

\mathsf{= 25 + 45 + 90 + 60 + 45 + 30 + 10 + 20 + 45 + 15 + 25}

\mathsf{= 70 + 90 + 60 + 45 + 30 + 10 + 20 + 45 + 15 + 25}

\mathsf{= 160 + 60 + 45 + 30 + 10 + 20 + 45 + 15 + 25}

\mathsf{= 220 + 45 + 30 + 10 + 20 + 45 + 15 + 25}

\mathsf{= 265 + 30 + 10 + 20 + 45 + 15 + 25}

\mathsf{= 295 + 10 + 20 + 45 + 15 + 25}

\mathsf{= 305 + 20 + 45 + 15 + 25}

\mathsf{= 325 + 45 + 15 + 25}

\mathsf{= 370 + 15 + 25}

\mathsf{= 385 + 25}

\mathsf{= \bf 410}

\mathsf{= \dfrac{410}{12}}

\mathsf{\dfrac{410}{12} \approx \bf 34.1667}

\boxed{\boxed{\huge\textsf{Thus, your ANSWER is: \boxed{\mathsf{ \bf \underline{{\star 34.1667\star}}}}}}}}\huge\checkmark

\huge\textsf{Good luck on your assignment \& enjoy your day!}

~\frak{Amphitrite1040:)}

6 0
3 years ago
Read 2 more answers
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