A radar operator on a ship discovers a large sunken vessel lying parallel to the ocean surface, 150 m

1 Given the data 21, 13, 13, 37, 13, 23, 25, 15:

insens350 [35]

A. the outlier is 37
B. the mean with the outlier is 20
C. the mean without the outlier is 17.5

3 0

3 years ago

15 Ellen is at a hydroplane boat show. Starting from 1.0 miles away, the boat drives towards her and then passes her. The boat p

german

D = d + Vt,
d = 1.0 miles,
V = -70 miles per hour,(- because its coming towards you)
And about D
Imagine you're standing at the centre of the street. a car is coming towards you. You can easily see that car will be X miles away from you two times. When Driving towards you And When Driving away from you. So
we have two D. + 0.15 miles and -0.15 miles.
0.15= 1 - 70t
-0.15=1-70t
From here
t equals
either
0.85/70 hours which is 0.85/70*3600 second about 43.7 sec.
or
1.15/70 hours or 1.15*3600/70 about 59 sec.

6 0

3 years ago

Find z when x = 4 and y = 9 if z varies inversely with the product of x and y when x = 2, y = 4, and z = 0.5

Vera_Pavlovna [14]

8=9*87-8 9+9 765 it could be OK

8 0

3 years ago

There are three boxes: one with two golden coins, one with two silver coins, and one with one golden coin and one silver coin. A

krek1111 [17]

Answer:

Step-by-step explanation:

From Bayes' theorem is stated mathematically as the following equation:[2]

{\displaystyle P(A\mid B)={\frac {P(B\mid A)\,P(A)}{P(B)}},}

where A and B are events and P(B) ≠ 0.

P(A) and P(B) are the probabilities of observing A and B without regard to each other.

P(A | B), a conditional probability, is the probability of observing event A given that B is true.

P(B | A) is the probability of observing event B given that A is true.

At this point, go through the attached file before you continue with part B.

Part B)

P(silver) = P(silver from SS)+P(silver from GS)

note P(SS)=P(GG)=P(GS) = 1/3

P(silver from SS) = 1

P(silver from GS) = 1/2

hence

P(Silver from SS) = 1/3

P(Silver from GS) = 1/3 *1/2

P(Silver) = 1/3*1+1/3*1/2

required probability = P(Silver from SS)/P(Silver) = 2/3

4 0

4 years ago

Suppose we have two thermometers. One thermometer is very precise but is delicate and heavy (X). We have another thermometer tha

Temka [501]

Answer:

a)H0: β1= 1 and Ha: β1 ≠1

b) The test statistic is____.t= b- β/ sb

c) The p-value is____.0.999144

(d) Therefore, we can conclude that:_____. there is not enough evidence to reject the null hypothesis.

2) The data provides no evidence at the 0.1 significance level that these thermometers are not consistent.

Step-by-step explanation:

The null and alternate hypotheses are

H0: β1= 1 and Ha: β1 ≠1

The significance level is alpha= 0.1 but ∝/2 =0.1/2= 0.05 for two tailed test.

The critical region at t ∝/2 (29)= t ≤ - 1.699 , t ≥ 1.699

The test statistic is

t= b- β/ sb

which has t distribution with υ= 31-2= 29 degrees of freedom.

Calculations

Ŷ = a +bX

b = SPxy /SS x= Σ(xi-x`)(yi-y`)/Σ(xi-x`)²

b = 39671.6/39680= 0.9998

a = y` - bx`

x`= 60

y`= 59.989

a = 59.989 -0.9998*60 = 0.001734

Syx²= Σ( yi -y`)²/ n-2= 39663.3857/ 29=1367.68965

Syx= 36.9822

Sb= Syx/ √∑ (x-x`)²

Sb= 36.9823/ √39680

Sb= 36.9823/ 199.984

Sb= 0.18493

x-x` y-y` (x-x`)² (x-x`)(y-y`)

-60 -59.969 3600 3598.1419

-56 -55.999 3136 3135.9458

-52 -52.079 2704 2708.1097

-48 -47.959 2304 2302.0335

-44 -43.899 1936 1931.5574

-40 -39.989 1600 1599.5613

-36 -36.009 1296 1296.3252

-32 -31.899 1024 1020.769

-28 -28.049 784 785.3729

-24 -23.959 576 575.0168

-20 -19.989 400 399.7806

-16 -15.939 256 255.0245

-12 -12.039 144 144.4684

-8 -7.989 64 63.9123

-4 -4.119 16 16.4761

0 -0.08903 0 0

4 3.921 16 15.6839

8 7.961 64 63.6877

12 12.121 144 145.4516

16 16.031 256 256.4955

20 20.021 400 400.4194

24 24.111 576 578.6632

28 28.071 784 785.9871

32 31.751 1024 1016.031

36 36.031 1296 1297.1148

40 39.961 1600 1598.4387

44 43.881 1936 1930.7626

48 48.021 2304 2305.0065

52 52.001 2704 2704.0503

56 56.051 3136 3138.8542

Putting the values

The test statistic is

t= b- β/ sb

t= 0.9998-1/ 0.18493 ( β=1 given)

t=-0.0010815

Since the calculated t=-0.0010815 does not lie in the critical region t ∝/2 (29)= t ≤ - 1.699 , t ≥ 1.699 we conclude that these thermometers seem to be measure temperatures.

The p-value is ≈ 0.999144

there is not enough evidence to reject the null hypothesis.

7 0

3 years ago