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Luda [366]
3 years ago
7

I need help!! Please help!!

Mathematics
1 answer:
Lady bird [3.3K]3 years ago
3 0
If she got a 100 on her last test the highest her average would be is an 87%
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Please help with my straight lines!!
Mrrafil [7]
Where is the question
8 0
3 years ago
HURRY
lesantik [10]
I think it is the e first one A because if the graph shows that it’s in one so I think the if you but it in one I think that the gragh also has to show 9
6 0
2 years ago
A manufacturing process outputs parts having a normal distribution with a mean of 30 cm and standard deviation of 2 cm. From a p
Arisa [49]

Answer:

68% of the sample can be expected to fall between 28 and 32 cm

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 30

Standard deviation = 2

What proportion of the sample can be expected to fall between 28 and 32 cm

28 = 30-2

28 is one standard deviation below the mean

32 = 30 + 2

32 is one standard deviation above the mean

By the Empirical Rule, 68% of the sample can be expected to fall between 28 and 32 cm

7 0
3 years ago
6.
dimulka [17.4K]

So car A travels Distance = 60 km/hr*(1+x)

Car B travels Distance = 75 km/hr(x)

Where x is the time in hours.

Those two equations are equal when on overtakes the other

so:

60 +60x = 75x

60=15x

x=4, but my expression is written to count from when car B commenced travel. So total time is 5 hours from the car A setting off.

3 0
2 years ago
Read 2 more answers
Evaluate the summation of negative 2 n minus 3, from n equals 2 to 10..
Alisiya [41]
\bf \sum\limits_{n=2}^{10}~-2n-3

so, let's first change the bounds from 1 to 9, so we'll be dropping the indices by 1, in order to do the variable "n".

however, let's use a variable hmmm say "k", where "k" is the value of "n" beginning at 1, thus k = n - 1.

however, if k = n - 1, then k + 1 = n.  So let's use those fellows then,

\bf \sum\limits_{n=2}^{10}~-2n-3\qquad 
\begin{cases}
k=n-1\\
k+1=n
\end{cases}\implies \sum\limits_{k=1}^{9}~-2\stackrel{n}{(k+1)}-3
\\\\\\
\sum\limits_{k=1}^{9}~-2k-2-3\implies \sum\limits_{k=1}^{9}~-2k-5\implies \sum\limits_{k=1}^{9}~-2k-\sum\limits_{k=1}^{9}~5
\\\\\\
-2\sum\limits_{k=1}^{9}~k-\sum\limits_{k=1}^{9}~5\implies -2\left( \cfrac{9(9+1)}{2} \right)\qquad -\qquad (9\cdot 5)
\\\\\\
-2(45)-45\implies -90-45\implies -135
5 0
3 years ago
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