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eimsori [14]
3 years ago
15

Is a whole number larger than a decimal? My problem is 2 _ 2.3

Mathematics
1 answer:
algol133 years ago
6 0
2.3 is larger than 2
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Let the sequence {an} be defined so that a1 = 1 and each succeeding term is found by adding 3 to the one before it. Find the for
Sauron [17]

Answer:

C

Step-by-step explanation:

The sequence is arithmetic with n th term

a_n} = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Here a₁ = 1 and d = 3, thus

a_{n} = 1 + 3(n - 1) = 1 + 3n - 3 = 3n - 2 → C

6 0
3 years ago
I need the answer this is hard
vlada-n [284]

it takes the robot 32/3 hour(s) to complete one task which in decimal form would be something like 10.6667.

3 0
2 years ago
The home run percentage is the number of home runs per 100 times at bat. A random sample of 43 professional baseball players gav
Andru [333]

Step-by-step explanation:

(a) Yes, if you enter all 43 values into your calculator, you calculator should report:

xbar = 2.293

s = 1.401

(b)

Note: Most professors say that is sigma = the population standard deviation is unknown (as it is unknown here), you should construct a t-confidence interval.

xbar +/- t * s / sqrt(n)

2.293 - 1.684 * 1.401 / sqrt(43) = 1.933

2.293 - 1.684 * 1.401 / sqrt(43) = 2.653

Answer: (1.933, 2.653)

Note: To find the t-value that allows us to be 90% confident, go across from df = 43-1 = 42 (round down to 40 to be conservative since 42 in not in the table) and down from (1-.90)/2 = .05 or up from 90% depending on your t-table. So, the t-critical value is 1.684.

Note: If you can use the TI-83/84, it will construct the following CI using df = 42 (ie t = 1.681).

2.293 +/- 1.681 * 1.401 / sqrt(43)

(1.934, 2.652)

Note: Some professors want you to construct a z-CI when the sample size is large. If your professor says this, the correct 90% CI is:

2.293 +/- 1.645 * 1.401 / sqrt(43)

(1.942, 2.644)

Note: To find the z-value that allows us to be 90% confident, (1) using the z-table, look up (1-.90)/2 = .05 inside the z-table, or (2) using the t-table, go across from infinity df (= z-values) and down from .05 or up from 90% depending on your t-table. Either way, the z-critical value is 1.645.

(c)

Note: Again, most professors say that is sigma = the population standard deviation is unknown (as it is unknown here), you should construct a t-confidence interval.

xbar +/- t * s / sqrt(n)

2.293 - 2.704 * 1.401 / sqrt(43) = 1.715

2.293 - 2.704 * 1.401 / sqrt(43) = 2.871

Answer: (1.715, 2.871)

Note: To find the t-value that allows us to be 99% confident, go across from df = 43-1 = 42 (round down to 40 to be conservative since 42 in not in the table) and down from (1-.99)/2 = .005 or up from 99% depending on your t-table. So, the t-critical value is 2.704.

Note: If you can use the TI-83/84, it will construct the following CI using df = 42 (ie t = 2.698).

2.293 +/- 2.698 * 1.401 / sqrt(43)

(1.717, 2.869)

Note: Again, some professors want you to construct a z-CI when the sample size is large. If your professor says this, the correct 99% CI is:

2.293 +/- 2.576 * 1.401 / sqrt(43)

(1.742, 2.843)

Note: To find the z-value that allows us to be 99% confident, (1) using the z-table, look up (1-.99)/2 = .005 inside the z-table, or (2) using the t-table, go across from infinity df (= z-values) and down from .005 or up from 99% depending on your t-table. Either way, the z-critical value is 2.576.

(d)

Tim Huelett 2.5

Since 2.5 falls between (1.715, 2.871), we see that Tim Huelett falls in the 99% CI range. So, his home run percentage is NOT significantly different than the population average.

Herb Hunter 2.0

Since 2.0 falls between (1.715, 2.871), we see that Herb Hunter falls in the 99% CI range. So, his home run percentage is NOT significantly different than the population average.

Jackie Jensen 3.8.

Since 3.8 falls above (1.715, 2.871), we see that Jackie Jensen falls in the 99% CI range. So, his home run percentage IS significantly GREATER than the population average.

(e)

Because of the Central Limit Theorem (CLT), since our sample size is large, we do NOT have to make the normality assumption since the CLT tells us that the sampling distribution of xbar will be approximatley normal even if the underlying population distribution is not.

6 0
2 years ago
Consider functions f and g below.
Pani-rosa [81]

Answer:

D

Step-by-step explanation:

f(x) - g(x) = -9x^2 - 7x + 12 - (3x^2 - 4x - 15)  Remove the brackets

f(x) - g(x) = -9x^2 - 7x  + 12 - 3x^2 + 4x + 15

f(x) - g(x) = -9x^2 - 3x^2 - 7x + 4x + 12 + 15

f(x) - g(x) = -12x^2- 3x + 27

Answer D

3 0
3 years ago
Read 2 more answers
Select the correct answer from each drop-down menu.
serious [3.7K]

Answer:

Center: (4,8)

Radius: 2.5

Equation: (x-4)^2+(y-8)^2=6.25

Step-by-step explanation:

It was given that; the endpoints of the longest chord on a circle are (4, 5.5) and (4, 10.5).

Note that the longest chord is the diameter;

The midpoint of the ends of the diameter gives us the center;

Use the midpoint formula;

(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} )

The center is at; (\frac{4+4)}{2} ,\frac{5.5+10.5}{2}=(4,8)

To find the radius, use the distance formula to find the distance from the center to one of the endpoints.

The distance formula is;

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

r=\sqrt{(4-4)^2+(10.5-8)^2}

r=\sqrt{0^2+(2.5)^2}

r=\sqrt{0^2+(2.5)^2}=2.5

The equation of the circle in standard form is given by;

(x-h)^2+(y-k)^2=r^2

We substitute the center and the radius into the formula to get;

(x-4)^2+(y-8)^2=2.5^2

(x-4)^2+(y-8)^2=6.25

6 0
3 years ago
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