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scoray [572]
3 years ago
8

Solve plz help 2/3 + x = 3

Mathematics
2 answers:
Sergeu [11.5K]3 years ago
8 0

Answer:

x=7/3 or 2 and 1/3

Step-by-step explanation:

subtract the 2/3 from 3 and you get 7/3

Darya [45]3 years ago
4 0

Answer:

7/3

Step-by-step explanation:

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Is the radio is not 30??
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3 years ago
SERIOUS ANSWERS ONLY WILL GIVE BRAINLIEST
tensa zangetsu [6.8K]

Hello,

Part A:

f(x)=4x^2+8x-5\\=4(x^2+2x)-5\\=4(x^2+2x+1-1)-5\\=4(x+1)^2-9\\=(2(x+1)-3)(2(x+1)+3)\\=(2x-1)(2x+5)\\x-intercepts\ are\ x=\frac{1}{2} \ and\ x=-\frac{5}{2} \\

Part B:

x² coefficient is 4 >0 thus a minimun

as y=4(x+1)²-9 : vertex is (-1,-9)

Proof: see picture

Sorry for Part c: I don't know

In France, we make an array of (x,f(x)) and then plot the severals points.

4 0
3 years ago
Can u answer dis?<br><br> will mark brainliest
Grace [21]
The correct answer is B
325-45=280
8x35=280
3 0
2 years ago
A telephone exchange operator assumes that 8% of the phone calls are wrong numbers. If the operator is right, what is the probab
Vera_Pavlovna [14]

Answer:

The probability that the sample proportion differ from the population proportion by greater than 3% is 0.0241.

Step-by-step explanation:

Let <em>X</em> = number of phone calls that are wrong numbers.

The proportion of phone calls that are wrong numbers is, <em>p</em> = 0.08.

A sample of<em> </em><em>n</em> = 421 phone calls is selected to determine the proportion of wrong numbers in this sample.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of a Binomial distribution is:

P(X=x)={421\choose x}0.08^{x}(1-0.08)^{421-x}

Now, for the sample proportion to differ from the population proportion by 3% the value of the sample proportion should be:

\hat p-p=0.03\\\hat p-0.08=0.03\\\hat p=0.11                            \hat p-p=-0.03\\\hat p-0.08=-0.03\\\hat p=0.05

So when the sample proportion is less than 5% or greater than 11% the difference between the sample proportion and population proportion will be greater than 3%.

  • If sample proportion is 5% then the value of <em>X</em> is,

        X=np=421\times 0.05=21.05\approx21

        Compute the value of P (X ≤ 21) as follows:

       P(X\leq 21)=\sum\limits^{21}_{x=0}{{421\choose x}0.08^{x}(1-0.08)^{421-x}}=0.0106

  • If the sample proportion is 11% then the value of <em>X</em> is,

        X=np=421\times 0.11=46.31\approx47

        Compute the value of P (X ≥ 47) as follows:

       P(X\geq 47)=\sum\limits^{471}_{x=47}{{421\choose x}0.08^{x}(1-0.08)^{421-x}}=0.0135

Then the probability that the sample proportion differ from the population proportion by greater than 3% is:

P(\hat p-p>0.03)=P(X\leq 21)+P(X\geq 47)=0.0106+0.0135=0.0241

Thus, the probability that the sample proportion differ from the population proportion by greater than 3% is 0.0241.

7 0
3 years ago
Decide if the following biconditional statement
yawa3891 [41]

Answer:

True

Step-by-step explanation:

Pretty sure it's true.

7 0
3 years ago
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