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Natali5045456 [20]
3 years ago
11

Each unit cost 14p how much do 942 units

Mathematics
2 answers:
slava [35]3 years ago
6 0
14×942
= 13188is the answer
horrorfan [7]3 years ago
5 0
You would really just have to multiply to find the answer to that. 942 × 14 = 13,188. Meaning, 942 units would cost 13,188.
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Simplify the given polynomial expressions, and determine the degree and number of terms in each expression
Sunny_sXe [5.5K]

Answer: What is the "given polynomial expressions?"

6 0
3 years ago
Will mark BRAINLIEST to correct answer! Please help, this has to be correct. Thanks!
pogonyaev
The answer would be 314.16 square cm
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4 0
3 years ago
Read 2 more answers
Need help help on this question
balu736 [363]

Answer:70

Step-by-step explanation:

step 1:add up the angles you know

step 2:subtract the answer (220) from the total angle measure of a square which is 360 then you get 140

step 3: divide 140 by 2 because the two angles on the sides are equivalent and you get 70

6 0
3 years ago
If ab= 7x-6 and bc= 12-2x. Then what does ac equal?
ra1l [238]
Ab = (7x-6) →   b = (7x-6)/a   (1)

bc = (12-2x) → b = (12-2x)/c   (2)

Since (1) = (2) → (7x-6)/a = (12-2x)/c OR  a/c = (7x-6)/(12-2x)  (3)

Multiply  both numerators of (3) by the SQUARE of their  respective denominators;

(a*c²)/c = (7x-6)(12-2x)²/(12-2x)
Now simplify:

ac = (7x-6)(12x-2x)  or ac = -14x² + 96x - 72


4 0
3 years ago
25% of American households have only dogs (one or more dogs) 15% of American households have only cats (one or more cats) 10% of
sergeinik [125]

Answer:

a) P=0.2503

b) P=0.2759

c) P=0.3874

d) P=0.2051

Step-by-step explanation:

We have this information:

25% of American households have only dogs (one or more dogs)

15% of American households have only cats (one or more cats)

10% of American households have dogs and cats (one or more of each)

50% of American households do not have any dogs or cats.

The sample is n=10

a) Probability that exactly 3 have only dogs (p=0.25)

P(x=3)=\binom{10}{3}0.25^30.75^7=120*0.01563*0.13348=0.25028

b) Probability that exactly 2 has only cats (p=0.15)

P(x=2)=\binom{10}{2}0.15^20.85^8=45*0.0225*0.27249=0.2759

c) Probability that exactly 1 has cats and dogs (p=0.1)

P(x=1)=\binom{10}{1}0.10^10.90^0=10*0.1*0.38742=0.38742

d) Probability that exactly 4 has neither cats or dogs (p=0.5)

P(x=4)=\binom{10}{4}0.50^40.50^6=210*0.0625*0.01563=0.20508

8 0
3 years ago
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