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4 years ago

Please help! I’ll give BRAINLIEST :))

Mathematics

1 answer:

blondinia [14]4 years ago

6 0

Answer:

yes

Step-by-step explanation:

well converse is

If q, then p

and inverse is

If no p, then not q

butttttt lets put this in human terms

If I am allergic to donuts, then I don't like to eat donuts.

and the inverse is

If I don't like to eat donuts, then I am allergic to donuts.

It is true no matter what! However, there could also be other reasons you don't liek donuts but nonthless this is one true reason that exists bc converse.

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Item 14

dsp73

Answer:

13/21

Step-by-step explanation:

5 0

3 years ago

32 lb 12 ounces equals in pounds

Radda [10]

Answer:

32 3/4 lb

Step-by-step explanation:

32 lb 12 oz = ? lb

Rewrite 12 oz as (12/16) lb.

Then:

32 lb 12 oz = 32 lb + (3/4) lb, or 32 3/4 lb

4 0

3 years ago

You have a new mobile phone. The phone costs £140 and then you pay for a one year contract at £15 per month. How much do you pay

netineya [11]

It is the amount you pay each month times 12
15x12=£180

6 0

4 years ago

Least common multiple of 70, 60, and 50

goldenfox [79]

70 = 7 x 2 x 5
60 = 2² x 3 x 5
50 = 2 x 5²

LCM = 2 x 3 x 5 x 7 = 210

Answer: LCM = 210

3 0

4 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