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Murrr4er [49]
3 years ago
7

Help me on these mathematics questions please?

Mathematics
2 answers:
Klio2033 [76]3 years ago
7 0
Think i got it 
E= 75
F= 90
G= 25 
CaHeK987 [17]3 years ago
7 0

Answer:

x=61

E=65

F=90

G=25

Step-by-step explanation:


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How do you do this question?
quester [9]

Step-by-step explanation:

F(x) = ∫ₐˣ t⁷ dt

F(x) is the area under f(t) between t=a and t=x.  When x=a, the width of the interval is 0, so the area is zero.

F(6) = 0, so a = 6.

F(x) = ∫₆ˣ t⁷ dt

F(6) = ∫₆⁶ t⁷ dt

F(6) = 0

3 0
3 years ago
A boat goes 24 miles at a constant speed in 4 hours against a river current.
Ne4ueva [31]

Answer:

8 miles per hour.

Step-by-step explanation:

24miles in 4 hours = 24/4=6 miles per hour.

24 miles in 3 hours = 24/3=8 miles per hour. The boat going with the current of the river is traveling at 8 miles per hour.

7 0
3 years ago
Please Do These Three Equations And Show Work
german

2) Add 0.1 to both sides. v/2.2=7.5

Then multiply by 2.2 on both sides.

v=16.5

4) Subtract 1.9 from both sides. -1.3g=-13

Divide both sides by -1.3

g=10

6) Add 1.4 to both sides. -12.9=-3d

Divide -3 on both sides.

d=4.3

3 0
3 years ago
The graph of an exponential function is given. Which of the following is the correct equation of the function?
katen-ka-za [31]

Answer:

If one of the data points has the form  

(

0

,

a

)

, then a is the initial value. Using a, substitute the second point into the equation  

f

(

x

)

=

a

(

b

)

x

, and solve for b.

If neither of the data points have the form  

(

0

,

a

)

, substitute both points into two equations with the form  

f

(

x

)

=

a

(

b

)

x

. Solve the resulting system of two equations in two unknowns to find a and b.

Using the a and b found in the steps above, write the exponential function in the form  

f

(

x

)

=

a

(

b

)

x

.

EXAMPLE 3: WRITING AN EXPONENTIAL MODEL WHEN THE INITIAL VALUE IS KNOWN

In 2006, 80 deer were introduced into a wildlife refuge. By 2012, the population had grown to 180 deer. The population was growing exponentially. Write an algebraic function N(t) representing the population N of deer over time t.

SOLUTION

We let our independent variable t be the number of years after 2006. Thus, the information given in the problem can be written as input-output pairs: (0, 80) and (6, 180). Notice that by choosing our input variable to be measured as years after 2006, we have given ourselves the initial value for the function, a = 80. We can now substitute the second point into the equation  

N

(

t

)

=

80

b

t

to find b:

⎧

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎨

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎩

N

(

t

)

=

80

b

t

180

=

80

b

6

Substitute using point  

(

6

,

180

)

.

9

4

=

b

6

Divide and write in lowest terms

.

b

=

(

9

4

)

1

6

Isolate  

b

using properties of exponents

.

b

≈

1.1447

Round to 4 decimal places

.

NOTE: Unless otherwise stated, do not round any intermediate calculations. Then round the final answer to four places for the remainder of this section.

The exponential model for the population of deer is  

N

(

t

)

=

80

(

1.1447

)

t

. (Note that this exponential function models short-term growth. As the inputs gets large, the output will get increasingly larger, so much so that the model may not be useful in the long term.)

We can graph our model to observe the population growth of deer in the refuge over time. Notice that the graph below passes through the initial points given in the problem,  

(

0

,

8

0

)

and  

(

6

,

18

0

)

. We can also see that the domain for the function is  

[

0

,

∞

)

, and the range for the function is  

[

80

,

∞

)

.

Graph of the exponential function, N(t) = 80(1.1447)^t, with labeled points at (0, 80) and (6, 180).If one of the data points has the form  

(

0

,

a

)

, then a is the initial value. Using a, substitute the second point into the equation  

f

(

x

)

=

a

(

b

)

x

, and solve for b.

If neither of the data points have the form  

(

0

,

a

)

, substitute both points into two equations with the form  

f

(

x

)

=

a

(

b

)

x

. Solve the resulting system of two equations in two unknowns to find a and b.

Using the a and b found in the steps above, write the exponential function in the form  

f

(

x

)

=

a

(

b

)

x

.

EXAMPLE 3: WRITING AN EXPONENTIAL MODEL WHEN THE INITIAL VALUE IS KNOWN

In 2006, 80 deer were introduced into a wildlife refuge. By 2012, the population had grown to 180 deer. The population was growing exponentially. Write an algebraic function N(t) representing the population N of deer over time t.

SOLUTION

We let our independent variable t be the number of years after 2006. Thus, the information given in the problem can be written as input-output pairs: (0, 80) and (6, 180). Notice that by choosing our input variable to be measured as years after 2006, we have given ourselves the initial value for the function, a = 80. We can now substitute the second point into the equation  

N

(

t

)

=

80

b

t

to find b:

⎧

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎨

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎩

N

(

t

)

=

80

b

t

180

=

80

b

6

Substitute using point  

(

6

,

180

)

.

9

4

=

b

6

Divide and write in lowest terms

.

b

=

(

9

4

)

1

6

Isolate  

b

using properties of exponents

.

b

≈

1.1447

Round to 4 decimal places

.

NOTE: Unless otherwise stated, do not round any intermediate calculations. Then round the final answer to four places for the remainder of this section.

The exponential model for the population of deer is  

N

(

t

)

=

80

(

1.1447

)

t

. (Note that this exponential function models short-term growth. As the inputs gets large, the output will get increasingly larger, so much so that the model may not be useful in the long term.)

We can graph our model to observe the population growth of deer in the refuge over time. Notice that the graph below passes through the initial points given in the problem,  

(

0

,

8

0

)

and  

(

6

,

18

0

)

. We can also see that the domain for the function is  

[

0

,

∞

)

, and the range for the function is  

[

80

,

∞

)

.

Graph of the exponential function, N(t) = 80(1.1447)^t, with labeled points at (0, 80) and (6, 180).

Step-by-step explanation:

4 0
3 years ago
Babies born after 40 weeks gestation have a mean length of 52.2 centimeters (about 20.6 inches). Babies born one month early hav
Sveta_85 [38]

Answer:

a) Z = -2.88

b) Z = -0.96

c) 40 weeks gestation babies

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

a. Find the standardized score (z-score), relative to all U.S. births, for a baby with a birth length of 45 cm.

Here, we use \mu = 52.2, \sigma = 2.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{45 - 52.2}{2.5}

Z = -2.88

b. Find the standardized score of a birth length of 45 cm. for babies born one month early, using 47.4 as the mean.

Here, we use \mu = 47.4, \sigma = 2.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{45 - 47.4}{2.5}

Z = -0.96

c. For which group is a birth length of 45 cm more common?

For each group, the probability is 1 subtracted by the pvalue of Z.

Z = -2.88 has a lower pvalue than Z = -0.96, so for Z = -2.88 the probability 1 - pvalue of Z will be greater. This means that for 40 weeks gestation babies a birth length of 45 cm is more common.

3 0
3 years ago
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