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Vesna [10]
3 years ago
12

Express 2 1/2kg as grams

Mathematics
1 answer:
enyata [817]3 years ago
8 0
The answer is 2 1/2kg = <span>2500 g</span>
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What is the constant term of -3x^4-7x+2
emmasim [6.3K]

Answer:

The equation written in the standard form is written as

ax^4 + bx+c

where a and b are the coefficients of the second-order and first-order term, while c is the constant term.

The equation we have in this problem is

By comparing (1) with (2), we immediately see that

A = -3x^4

B = 7x

C = 2

Step-by-step explanation:

Hope this helps :)

3 0
3 years ago
HELP ME
BartSMP [9]

Answer:

7.07?

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
e 400 trials in which the assistant maintained a neutral expression, the driver stopped in 229 out of the 400 trials. Find a 95%
Fittoniya [83]

Answer:

Step-by-step explanation:

n = 400

Proportion p = 229/400= 0.5725

For 95% confidence interval we use Z value as 1.96

Std error = 0.025

margin of error = 1.96*0.025

Confidence interval 95% = 0.5725±Margin of error

= (0.524, 0.621)

b) When smiled x becomes 277

p = 0.6925

Std error = 0.023

Margin of error= 1.96*0.023

Confidence interval = (0.647, 0.738)

Smiling increases the chances of stopping since mean and conidence interval bounds are showing increasing trend.

6 0
3 years ago
In a certain community, 36 percent of the families own a dog and 22 percent of the families that own a dog also own a cat. In ad
zysi [14]

Answer:  a) 0.0792   b) 0.264

Step-by-step explanation:

Let Event D = Families own a dog .

Event C = families own a cat .

Given : Probability that families own a dog : P(D)=0.36

Probability that families own a dog also own a cat : P(C|D)=0.22

Probability that families own a cat : P(C)= 0.30

a) Formula to find conditional probability :

P(B|A)=\dfrac{P(A\cap B)}{P(A)}\\\\\Rightarrow P(A\cap B)=P(B|A)\times P(A)   (1)

Similarly ,

P(C\cap D)=P(C|D)\times P(D)\\\\=0.22\times0.36=0.0792

Hence, the probability that a randomly selected family owns both a dog and a cat : 0.0792

b) Again, using (2)

P(D|C)=\dfrac{P(C\cap D)}{P(C)}\\\\=\dfrac{0.0792}{0.30}=0.264

Hence, the conditional probability that a randomly selected family owns a dog given that it owns a cat = 0.264

5 0
3 years ago
!! please help me i’ll mark brainliest !!
S_A_V [24]

Answer:

1/2=(A+B)

area of triangle=1/2×base×height

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3 years ago
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