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3 years ago

The equation Y-3 = -2(x+5) is written in point-slope form. What is the y-intercept of the line?

Mathematics

1 answer:

gladu [14]3 years ago

7 0

Answer:

B,-7

Step-by-step explanation:

convert the formualt into slope intercept form by adding the 3 and distributing the -2

then you will get y=-2x-7

in slope intercept the B slot is the y intercept, and in this equation it is -7

You might be interested in

The number of bacteria in a refrigerated food product is given by N ( T ) = 27 T 2 − 180 T + 100 N(T)=27T2-180T+100, 7 < T &l

NeX [460]

Answer:

$N(T(t))=432t^2-374.4t-118.88$

The number of Bactria after 5.8 hours is 12242.

Step-by-step explanation:

The number of bacteria in a refrigerated food product is given by

$N(T)=27T^2-180T+100$

where, T is the temperature of the food.

When the food is removed from the refrigerator, then the temperature is given by

$T(t)=4t+1.6$

We need to find the composite function N(T(t)).

$N(T(t))=N(4t+1.6)$

$N(T(t))=27(4t+1.6)^2-180(4t+1.6)+100$

$N(T(t))=432t^2+345.6t+69.12-720t-288+100$

$N(T(t))=432t^2-374.4t-118.88$

where N(T(t)) is the number of bacteria after t hours.

Substitute t=5.8 in the above function.

$N(T(5.8))=432(5.8)^2-374.4(5.8)-118.88$

$N(T(5.8))=14532.48-2290.4$

$N(T(5.8))=12242.08$

$N(T(5.8))\approx 12242$

Therefore, the number of Bactria after 5.8 hours is 12242.

7 0

4 years ago

NEED ANSWER STAT (PLEASEEEEEEEE)

Naddika [18.5K]

Answer:

For pages the answer 17/4 or 4 and 1/4

For Minutes the answer is 4.5 or 4 and 1/2

Step-by-step explanation:

I hope this helps please mark brainliest

4 0

3 years ago

Suppose the number of children in a household has a binomial distribution with parameters n=12n=12 and p=50p=50%. Find the proba

nadya68 [22]

Answer:

a) 20.95% probability of a household having 2 or 5 children.

b) 7.29% probability of a household having 3 or fewer children.

c) 19.37% probability of a household having 8 or more children.

d) 19.37% probability of a household having fewer than 5 children.

e) 92.71% probability of a household having more than 3 children.

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

In this problem, we have that:

$n = 12, p = 0.5$

(a) 2 or 5 children

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.1934$

$p = P(X = 2) + P(X = 5) = 0.0161 + 0.1934 = 0.2095$

20.95% probability of a household having 2 or 5 children.

(b) 3 or fewer children

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002$

$P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0002 + 0.0029 + 0.0161 + 0.0537 = 0.0729$

7.29% probability of a household having 3 or fewer children.

(c) 8 or more children

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 8) = C_{12,8}.(0.5)^{8}.(0.5)^{4} = 0.1208$

$P(X = 9) = C_{12,9}.(0.5)^{9}.(0.5)^{3} = 0.0537$

$P(X = 10) = C_{12,10}.(0.5)^{10}.(0.5)^{2} = 0.0161$

$P(X = 11) = C_{12,11}.(0.5)^{11}.(0.5)^{1} = 0.0029$

$P(X = 12) = C_{12,12}.(0.5)^{12}.(0.5)^{0} = 0.0002$

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.1208 + 0.0537 + 0.0161 + 0.0029 + 0.0002 = 0.1937$

19.37% probability of a household having 8 or more children.

(d) fewer than 5 children

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002$

$P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537$

$P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.1208$

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0002 + 0.0029 + 0.0161 + 0.0537 + 0.1208 = 0.1937$

19.37% probability of a household having fewer than 5 children.

(e) more than 3 children

Either a household has 3 or fewer children, or it has more than 3. The sum of these probabilities is 100%.

From b)

7.29% probability of a household having 3 or fewer children.

p + 7.29 = 100

p = 92.71

92.71% probability of a household having more than 3 children.

5 0

3 years ago

Hector went shopping for a computer. At RST store, a computer originally prices at $955 had a price reduction of 40%. What was t

balandron [24]

Answer:

$573

Step-by-step explanation:

955* 0.60=573

10% of 955= 95.5

50% of 955=477.5

477.5=95.5=573

4 0

3 years ago

HELP pleasee and ty to whoever does!!

Margaret [11]

Answers:
26) 50 percent chance
27) 25 percent chance
28) About 25 times
29) About 20 times

6 0

3 years ago