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2 years ago

Subtracting 766 from 941 results in a difference of

Mathematics

2 answers:

SpyIntel [72]2 years ago

7 0

941 - 766 = 175
Therefore, A

weeeeeb [17]2 years ago

7 0

941-766 is 175.

You can also add 175 to 766 to get 941.

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A baker used 5 7/10 bags of flour one week. The same baker used 7 3?10 bags the next week. How many more bags of flour did the b

Elenna [48]

So first y do is is add 5+7 and add 3/10 + 7/10 equal 10/10 so 5+7 + 1 is 13 she used 13 bags of flour the second week she used 1 more than last week

8 0

2 years ago

Suppose the sediment density(g/cm) of a randomly selected specimenfrom a certain region is normally distributed with mean 2.65 a

antiseptic1488 [7]

Answer:

ai ) $P(\= X \le 3.0 ) =0.980$

aii) $P(2.65 \le \= X \le 3.00) = 0.480$

b ) $n = 32$

Step-by-step explanation:

From the question we are told that

The mean is $\mu = 2.65$

The standard deviation is $\sigma = 0.85$

Let the random sediment density be X

given that the sediment density is normally distributed it implies that

X N(2.65 , 0.85)

Now probability that the sample average is at 3.0 is mathematically represented as

$P(\= X \le 3.0 ) = P[\frac{\= X - \mu} {\frac{\sigma }{\sqrt{n} } } \le \frac{3.0 - \mu}{\frac{ \sigma}{\sqrt{n} } } ]$

Here n is the sample size = 25 and $\= X$ is the sample mean

Now Generally the Z-value is obtained using this formula

$Z = \frac{\= X - \mu} {\frac{\sigma }{\sqrt{n} } }$

Thus

$P(\= X \le 3.0 ) = P[Z \le \frac{3.0 - 2.35}{\frac{ 0.85}{\sqrt{25} } } ]$

$P(\= X \le 3.0 ) = P[Z \le 2.06 ]$

From the z-table the z-score is 0.980

Thus

$P(\= X \le 3.0 ) =0.980$

Now probability that the sample average is between 2.65 and 3.00 is mathematically evaluated as

$P(2.65 \le \= X \le 3.00) = P [\frac{2.65 - \mu }{ \frac{\sigma }{\sqrt{n} } } < \frac{\= X - \mu }{ \frac{\sigma }{\sqrt{n} } } < \frac{3.0 - \mu }{ \frac{\sigma }{\sqrt{n} } }]$

$P(2.65 \le \= X \le 3.00) = P [\frac{2.65 - 2.65 }{ \frac{0.85 }{\sqrt{25} } }$

$P(2.65 \le \= X \le 3.00) = P[0 < Z< 2.06]$

$P(2.65 \le \= X \le 3.00) = P(Z < 2.06) - P(Z$

From the z-table

$P(2.65 \le \= X \le 3.00) = 0.980 - 0.50$

$P(2.65 \le \= X \le 3.00) = 0.480$

Now from the question

$P(\= X \le 3.0 ) =0.99$

=> $P(\= X \le 3.0 ) = P[Z \le \frac{3.0 - 2.35}{\frac{ 0.85}{\sqrt{n} } } ] = 0.99$

Generally the critical value of z for a one tail test such as the one we are treating that is under the area 0.99 is $t_z = 2.33$ this is obtained from the critical value table