If there is 7018 mm, what is the metric unit for 7.018?

Vanyuwa [196]

Answer: To find a residual you must take the predicted value and subtract it from the measured value.

Step-by-step explanation:

7 0

3 years ago

It takes maria ten hours to pick up forty bushels of apples kayla can pick the same amount in 12 hours how long would it take th

bearhunter [10]

answer: 11 hours

Step-by-step explanation: if they work together and split the forty bushes to pick in half {divide by two} then they pick twenty bushes of apples each. So they work half less {divide by two} . So it Maria would need five hours while kayla works six { add together} which equals 11 hours

4 0

3 years ago

Alan bought a suit on sale for $208. This price was 35% less than the original price.

algol [13]

Answer:

280.8

Step-by-step explanation:

Since 100% is the full amount, and it is 35% off you add 35 to the 100. So you get 1.35. You finally multiply by 207 and get 280.8

7 0

3 years ago

Patients in a hospital are classified as surgical or medical. A record is kept of the number of times patients require nursing s

GREYUIT [131]

Answer:

Following are the solution to the given question:

Step-by-step explanation:

Given:

$N= 177\\\\a=46\\\\b=52\\\\c=36\\\\d=43$

$\to \chi _{1}^{2}=\frac{N(ad-bc)^{2}}{(a+c)(b+d)(a+b)(c+d)}$

$=\frac{177((46\times 43)-(52 \times 36))^{2}}{(98)(79)(82)(95)}\\\\=\frac{177((1978)-(1872))^{2}}{60310180}\\\\=\frac{177(106)^{2}}{60310180}\\\\=\frac{177(11236)}{60310180}\\\\=\frac{1988772}{60310180}\\\\= 0.032976$

P-value= $CHIDIST(0.032976,1) = 0.86.$

thus, the surgical-medical patients and Medicare are dependent.

5 0

3 years ago

If bolt thread length is normally distributed, what is the probability that the thread length of a randomly selected bolt is Wit

KatRina [158]

Answer:

a) 0.5762

b) 0.0214

c) 0.2718

Step-by-step explanation:

It is given that lengths of the bolt thread are normally distributed. So in order to find the required probability we can use the concept of z distribution and z scores.

Part a) Probability that length is within 0.8 SDs of the mean

We have to calculate the probability that the length of a bolt thread is within 0.8 standard deviations of the mean. Recall that a z- score tells us that how many standard deviations away a value is from the mean. So, indirectly we are given the z-scores here.

Within 0.8 SDs of the mean, means from a score of -0.8 to +0.8. i.e. we have to calculate:

P(-0.8 < z < 0.8)

We can find these values from the z table.

P(-0.8 < z < 0.8) = P(z < 0.8) - P(z < -0.8)

= 0.7881 - 0.2119

= 0.5762

Thus, the probability that the thread length of a randomly selected bolt is within 0.8 SDs of its mean value is 0.5762

Part b) Probability that length is farther than 2.3 SDs from the mean

As mentioned in previous part, 2.3 SDs means a z-score of 2.3.

2.3 Standard Deviations farther from the mean, means the probability that z scores is lesser than - 2.3 or greater than 2.3

i.e. we have to calculate:

P(z < -2.3 or z > 2.3)

According to the symmetry rules of z-distribution:

P(z < -2.3 or z > 2.3) = 1 - P(-2.3 < z < 2.3)

We can calculate P(-2.3 < z < 2.3) from the z-table, which comes out to be 0.9786. So,

P(z < -2.3 or z > 2.3) = 1 - 0.9786

= 0.0214

Thus, the probability that a bolt length is 2.3 SDs farther from the mean is 0.0214

Part c) Probability that length is between 1 and 2 SDs from the mean value

Between 1 and 2 SDs from the mean value can occur both above the mean and below the mean.

For above the mean: between 1 and 2 SDs means between the z scores 1 and 2

For below the mean: between 1 and 2 SDs means between the z scores -2 and -1

i.e. we have to find:

P( 1 < z < 2) + P(-2 < z < -1)

According to the symmetry rules of z distribution:

P( 1 < z < 2) + P(-2 < z < -1) = 2P(1 < z < 2)

We can calculate P(1 < z < 2) from the z tables, which comes out to be: 0.1359

So,

P( 1 < z < 2) + P(-2 < z < -1) = 2 x 0.1359

= 0.2718

Thus, the probability that the bolt length is between 1 and 2 SDs from its mean value is 0.2718

4 0

3 years ago