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Nadusha1986 [10]
3 years ago
10

Log2/3 9/4=x+3 ????????????????

Mathematics
2 answers:
irga5000 [103]3 years ago
8 0
log_{\frac{2}{3}}\frac{9}{4}=x^3\\
\\
\left(\frac{2}{3}\right)^x+3=\frac{9}{4}\\
\\
\left(\frac{2}{3}\right)^{x+3}=\left(\frac{2}{3}\right)^{-2}\\
\\
x+3=-2\\
\\
x=-2-3\\
\\
\boxed{x=-5}
Lilit [14]3 years ago
7 0
log_{ \frac{2}{3} } \frac{9}{4} =x+3\\\\log_{ \frac{2}{3} }( \frac{2}{3})^{-2} =x+3\\\\-2=x+3\\\\-x=3+2\\\\-x=5\ /\cdor(-1)\\\\x=-5
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general admission to the fair is 10.00. each ride costs 4.75. if lisa has 75.00 to spend at the fair, what method can be used to
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Answer:

well the answer is 13

Step-by-step explanation:

The way you can figure this out is subtract 10 from 75 to get 65 and then you divide 65 by 4.75 to get 13 with a huge decimal but you just have 13 because you cant go for part of a ride.

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The endpoints of a segment are (–5, –3) and (0, 2). What are the endpoints of the segment after it has been translated 4 units u
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4 units up means 4 is added to the y-coordinate

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4 years ago
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F(x)=-4x - 5 g(x)=3x - 2 Find( f - g )(x)
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#1 is 12x² -15x 

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On a recent trip to the convenience store, you picked up 4 gallons of milk, 6 bottles of water and 5 snack size bags of chips. Y
bearhunter [10]

Answer:

Milk $3.60

Water $1.80

Chips $0.90.

Step-by-step explanation:

If w = cost of 1 bottle of water, m = cost of 1 gal of milk and c cost of 1 bag of chips we have.

4m + 6w + 5c  = 29.7

Also we have w = 2c and m = 1.8 + w.

From the first equation we have c = w/2.

So substituting for c and m in the initial equation we have:

4(1.8 + w) + 6w + 5(w/2) = 29.7

7.2 + 4w + 6w + 2.5w = 29.7

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8 0
4 years ago
Mathematics achievement test scores for 300 students were found to have a mean and a variance equal to 600 and 3600, respectivel
Zina [86]

Answer:

(a) Approximately 205 students scored between 540 and 660.

(b) Approximately 287 students scored between 480 and 720.

Step-by-step explanation:

A mound-shaped distribution is a normal distribution since the shape of a normal curve is mound-shaped.

Let <em>X</em> = test score of a student.

It is provided that X\sim N(\mu = 600, \sigma^{2} = 3600).

(a)

The probability of scores between 540 and 660 as follows:

P(540\leq X\leq 660)=P(\frac{540-600}{\sqrt{3600} }\leq \frac{X-600}{\sqrt{3600} }\leq \frac{660-600}{\sqrt{3600} })\\=P(-1 \leq Z\leq 1)\\= P(Z\leq 1)-P(Z\leq -1)\\=0.8413-0.1587\\=0.6826

Use the standard normal table for the probabilities.

The number of students who scored between 540 and 660 is:

300 × 0.6826 = 204.78 ≈ 205

Thus, approximately 205 students scored between 540 and 660.

(b)

The probability of scores between 480 and 720 as follows:

P(480\leq X\leq 720)=P(\frac{480-600}{\sqrt{3600} }\leq \frac{X-600}{\sqrt{3600} }\leq \frac{720-600}{\sqrt{3600} })\\=P(-2 \leq Z\leq 2)\\= P(Z\leq 2)-P(Z\leq -2)\\=0.9772-0.0228\\=0.9544

Use the standard normal table for the probabilities.

The number of students who scored between 480 and 720 is:

300 × 0.9544 = 286.32 ≈ 287

Thus, approximately 287 students scored between 480 and 720.

3 0
3 years ago
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