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3 years ago

Please answer TIMED i will give 35 points.

Mathematics

1 answer:

NNADVOKAT [17]3 years ago

5 0

Answer:

32 units squared

Step-by-step explanation:

This is kind of a weird shape, so split it into smaller shapes, in this case two triangles and a rectangle.

Both of the triangles have a base of 8 and a height of 2. Multiply the base and the height together, then divide by 2. 8*2/2 equals 8, so the area of each triangle is 8.

For the rectangle, the length is 2, and the width is 8. Multiply the length and the width together to get 16.

Finally, add the areas of all those smaller shapes together to get the area of the whole shape, 32 units squared.

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What is equivalent to 48x-6+3y/3

DIA [1.3K]

Answer:

Step-by-step explanation:

We can divide 3 out of all three given terms:

3(16x - 2 + y/3)

is one possibility. Where are the answer choices that you were given?

Note: 48x-6+3y/3 is an expression in x and y, but is not an equation. Is that what you intended?

6 0

3 years ago

Consider the function f(x)=xln(x). Let Tn be the nth degree Taylor approximation of f(2) about x=1. Find: T1, T2, T3. find |R3|

Fynjy0 [20]

Answer:

R3 <= 0.083

Step-by-step explanation:

f(x)=xlnx,

The derivatives are as follows:

f'(x)=1+lnx,

f"(x)=1/x,

f"'(x)=-1/x²

f^(4)(x)=2/x³

Simialrly;

f(1) = 0,

f'(1) = 1,

f"(1) = 1,

f"'(1) = -1,

f^(4)(1) = 2

As such;

T1 = f(1) + f'(1)(x-1)

T1 = 0+1(x-1)

T1 = x - 1

T2 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2

T2 = 0+1(x-1)+1(x-1)^2

T2 = x-1+(x²-2x+1)/2

T2 = x²/2 - 1/2

T3 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2+f"'(1)/6(x-1)^3

T3 = 0+1(x-1)+1/2(x-1)^2-1/6(x-1)^3

T3 = 1/6 (-x^3 + 6 x^2 - 3 x - 2)

Thus, T1(2) = 2 - 1

T1(2) = 1

T2 (2) = 2²/2 - 1/2

T2 (2) = 3/2

T2 (2) = 1.5

T3(2) = 1/6 (-2^3 + 6 *2^2 - 3 *2 - 2)

T3(2) = 4/3

T3(2) = 1.333

Since;

f(2) = 2 × ln(2)

f(2) = 2×0.693147 =

f(2) = 1.386294

Since;

f(2) >T3; it is significant to posit that T3 is an underestimate of f(2).

Then; we have, R3 <= | f^(4)(c)/(4!)(x-1)^4 |,

Since;

f^(4)(x)=2/x^3, we have, |f^(4)(c)| <= 2

Finally;

R3 <= |2/(4!)(2-1)^4|

R3 <= | 2 / 24× 1 |

R3 <= 1/12

R3 <= 0.083

5 0

2 years ago

ILL GIVE U STUFFFFF PLZZZZZZZZZZZ ANSWER ITTTTTTTTT

belka [17]

Answer:

Slope: $-\frac{3}{4}$

Y-intercept: 399

Slope-intercept form: $y=-\frac{3}{4}x+399$

Step-by-step explanation:

1. The equation in Standard form is given in the problem:

$3y+4y=1596$

2. Then, to write the equation of the line in slope-intercept form $y=mx+b$ , where m is the slope and b is the y-intercept, you must solve for y as you can see below:

$4y=-3x+1596\\y=\frac{-3x+1596}{4}\\y=-\frac{3}{4}x+399$

3. Therefore, the slope and the y-intercept are:

$m=-\frac{3}{4}\\b=399$

5 0

3 years ago

Each hour hanna is able to run 4 kilometers. If she has already run 2 kilometers, what will her total distance be after 8 hours?

Vlada [557]

Answer:

34km

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

The U.S. government has devoted considerable funding to missile defense research over the past 20 years. The latest development

Bad White [126]

Answer:

a) Let the random variable X= "number of these tracks where SBIRS detects the object." in order to use the binomial probability distribution we need to satisfy some conditions:

1) Independence between the trials (satisfied)

2) A value of n fixed , for this case is 20 (satisfied)

3) Probability of success p =0.2 fixed (Satisfied)

So then we have all the conditions and we can assume that:

$X \sim Bin(n =20, p=0.8)$

b) $X \sim Bin(n =20, p=0.8)$

c) $P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

d) $P(X \geq 15) = P(X=15)+ .....+P(X=20)$

$P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

$P(X=16)=(20C16)(0.8)^{16} (1-0.8)^{20-16}=0.218$

$P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205$

$P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137$

$P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.058$

$P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.012$

$P(X\geq 15)=0.804208$

e) $E(X) = np = 20*0.8 = 16$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

Let the random variable X= "number of these tracks where SBIRS detects the object." in order to use the binomial probability distribution we need to satisfy some conditions:

1) Independence between the trials (satisfied)

2) A value of n fixed , for this case is 20 (satisfied)

3) Probability of success p =0.2 fixed (Satisfied)

So then we have all the conditions and we can assume that:

$X \sim Bin(n =20, p=0.8)$

Part b

$X \sim Bin(n =20, p=0.8)$

Part c

For this case we just need to replace into the mass function and we got:

$P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

Part d

For this case we want this probability: $P(X\geq 15)$

And we can solve this using the complement rule:

$P(X \geq 15) = P(X=15)+ .....+P(X=20)$

$P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

$P(X=16)=(20C16)(0.8)^{16} (1-0.8)^{20-16}=0.218$

$P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205$

$P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137$

$P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.058$

$P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.012$

$P(X\geq 15)=0.804208$

Part e

The expected value is given by:

$E(X) = np = 20*0.8 = 16$

5 0

2 years ago