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3 years ago

A triangle with three acute angles

Mathematics

1 answer:

kotegsom [21]3 years ago

6 0

Answer:

An Acute triangle

Step-by-step explanation:

An acute triangle (or acute-angled triangle) is a triangle with three acute angles (less than 90°).

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The diameters of bolts produced in a machine shop are normally distributed with a mean of 5.71 millimeters and a standard deviat

Harlamova29_29 [7]

Answer:

The bolts with diameter less than 5.57 millimeters and with diameter greater than 5.85 millimeters should be rejected.

Step-by-step explanation:

We have been given that the diameters of bolts produced in a machine shop are normally distributed with a mean of 5.71 millimeters and a standard deviation of 0.08 millimeters.

Let us find the sample score that corresponds to z-score of bottom 4%.

From normal distribution table we got z-score corresponding to bottom 4% is -1.75 and z-score corresponding to top 4% or data above 96% is 1.75.

Upon substituting these values in z-score formula we will get our sample scores (x) as:

$z=\frac{x-\mu}{\sigma}$

$-1.75=\frac{x-5.71}{0.08}$

$-1.75*0.08=x-5.71$

$-0.14=x-5.71$

$-0.14+5.71=x-5.71+5.71$

$5.57=x$

Therefore, the bolts with diameters less than 5.57 millimeters should be rejected.

Now let us find sample score corresponding to z-score of 1.75 as upper limit.

$1.75=\frac{x-5.71}{0.08}$

$1.75*0.08=x-5.71$

$0.14=x-5.71$

$0.14+5.71=x-5.71+5.71$

$5.85=x$

Therefore, the bolts with diameters greater than 5.85 millimeters should be rejected.

7 0

3 years ago

Which equation below represents a rate of change of -3?

Amanda [17]

The answer is C

Is shows a change of rate for the integer

Hope this helps

6 0

3 years ago

A box of 15 cloud markers cost $12. 70. A box of 42 cloud markers cost $31. 60. How much would a box of 50 cloud markers cost? W

dalvyx [7]

P=0.7N + 2.20
Hope this helps

3 0

2 years ago

Suppose that 15% of the fields in a given agricultural area are infested with the sweet potato whitefly. One hundred fields in t

hram777 [196]

Answer:

a. $\mu=15$

b. $\mu=7.8586 \ and \ \mu=22.1414$

c. Choice A- Based on limits above, it is unlikely that we would see x = 45, so it might be possible that the trials are not independent.

Step-by-step explanation:

a.Binomial distribution is defined by the expression

$P(X=k)=C_k^n.p^k.(1-p)^{n-k}$

Let n be the number of trials, $n=100$

and p be the probability of success, $p=15\%$

The mean of a binomial distribution is the probability x sample size.

$\mu=np=100\times0.15=15$

b.Limits within which p is approximately 95%

sd of a binomial distribution is given as: $\sigma=\sqrt npq\\q=1-p$

Therefore, $\sigma=\sqrt(100\times0.015\times0.85)=3.5707$

Use the empirical rule to find the limits. From the rule, approximately 95% of the observations are within to standard deviations from mean.

$sd_1=>\mu-2\sigma=15-2\times3.3507=7.8586\\sd_2=>\mu-2\sigma=15+2\times3.3507=22.1414$

Hence, approximately 95% of the observations are within 7.8586 and 22.1414 (areas of infestation).

c. $x=45$ is not within the limits in b above (7.8586,22.1414). X=45 appears to be a large area of infestation. A.Based on limits above, it is unlikely that we would see x = 45, so it might be possible that the trials are not independent.

4 0

4 years ago

Please helme guys.....

vesna_86 [32]

Sorry this took so long! I managed to do everything except for the parts involving standard deviation. (I tried but I couldn't quite figure it out.)

~Hope this helps!~

Download xlsx

3 0

3 years ago