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3 years ago

7 * ( -15 ): por favor, es urgente):

Mathematics

1 answer:

Mademuasel [1]3 years ago

8 0

Answer: From what I understand you want me to do 7*(-15)??? and the answer to that is -105.

Step-by-step explanation: All you have to do is 7*-15

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Find the distance between the points (−5,2) and (9.2)

forsale [732]

Step-by-step explanation:

Formula Used

By distance formula,

d (A,B)=√[(x₁ - x₂)² + (y₁ - y₂)²]

$\sf \longmapsto \: D = √(x2-x1)²+(y2-y1)²$

$\sf \longmapsto= √(9+5)²+(2-2)²$

$\sf \longmapsto= = √14²+09$

$\sf \red \longmapsto \boxed{= 14}$

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2 years ago

Given x=-3, y=6, and z=-4 xz/-2y

zimovet [89]

Answer:

-1

Step-by-step explanation:

sub in the values of x, y and z into the equation

you get:

(-3)(-4)/(-2)(6)

= 12/-12

= -1

8 0

3 years ago

Read 2 more answers

Please help me with this one

Whitepunk [10]

Answer:

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6 0

3 years ago

Please Answer Attachment Below Thanks

docker41 [41]

Answer:

Step-by-step explanation:

The interior angles of a heptagon add up to 5*180 = 900

x + 143 + 116 + 135 + 135 + 116 + 155 = 900 Add up the known angles

x + 800 = 900 Subtract 800 from both sides

x + 800 - 800 = 900 - 800 Combine

x = 100

8 0

3 years ago

From experience, it is known that on average 10% of welds performed by a particular welder are defective. if this welder is requ

bulgar [2K]

Binomial distribution can be used because the situation satisfies all the following conditions:1. Number of trials is known and remains constant (n)2. Each trial is Bernoulli (i.e. exactly two possible outcomes) (success/failure)3. Probability is known and remains constant throughout the trials (p)4. All trials are random and independent of the othersThe number of successes, x, is then given by $P(x)=C(n,x)p^x(1-p)^{n-x}$ where $C(n,x)=\frac{n!}{x!(n-x)!}$
Here we're given
p=0.10 [ success = defective ]
n=3

(a) x=0
$P(x)=C(n,x)p^x(1-p)^{n-x}$
$=C(3,0)0.1^0(1-0.1)^{3-0}$
$=1(1)(0.729)$
$=0.729$

(b) x=2
$P(x)=C(n,x)p^x(1-p)^{n-x}$
$=C(3,2)0.1^2(1-0.1)^{3-2}$
$=3(0.01)(0.9)$
$=0.027$

(c) x ≥ 2
$P(x)=\sum_{x=2}^3C(n,x)p^x(1-p)^{n-x}$
$=P(2)+P(3)$
$=C(n,2)p^2(1-p)^{n-2}+C(n,3)p^3(1-p)^{n-3}$
$=C(3,2)0.1^2(1-0.1)^{3-2}+C(3,3)0.1^3(1-0.1)^{3-3}$
$=3(0.01)(0.9)+1(0.001)1$
$=0.027+0.001$
$=0.028$

8 0

3 years ago