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3 years ago

7

During quarantine your goal was to do a total of 36 pull-ups and push ups in a minute. You end up doin 8 times as many push-ups

as pull-ups. How many pull-ups and push-ups did you complete in a minute?

2 answers:

Semmy [17]3 years ago

5 0

Answer:

36 times 8 = 288

Maslowich3 years ago

5 0

You did 36 pull ups, and if you’re doing 8 times as many push ups, that means you need to multiply the number of pull ups with 8 because you did 8 TIMES as many more push ups. so it would be 288 because 36x8 is 288

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3 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

So

$t_z = \frac{3.0 - 2.35}{\frac{ 0.85}{\sqrt{n} } }$

$2.33 = \frac{3.0 - 2.35}{\frac{ 0.85}{\sqrt{n} } }$

=> $n = 32$

4 0

4 years ago

Ignore the filled in answer, just my marker. ANSWER ASAPP

KatRina [158]

Answer:

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Distance = square root of 116

Distance = 11

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Answer:

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4 years ago