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3 years ago

Answer this correctly and explain your answer I’ll give you brainalist + 10 points

Mathematics

1 answer:

nekit [7.7K]3 years ago

3 0

Answer: A

Step-by-step explanation:

In a triangle, any two sides, when added up need to be greater the last side

9 + 16 > 20

9 + 20 > 16

16 + 20 > 9

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What is the measure of angle TRV? 20° 50° 60° 130°

Svetllana [295]

Answer: 130

Step-by-step explanation:

8 0

3 years ago

What is R+G=20<br> 7R+3G=9

Troyanec [42]

Step-by-step explanation:

hope I understand your qyestion

Equation 1 : R+G=20

Equation 2 : 7R+3G=9

Multiply Equation by 3, it becomes

3R+3G=60 ...... Equation 3

Now Equation 3 minus Equation 2

-4R = 51

R = - 51/4

Sub R into Equation 1, you can find G

4 0

3 years ago

Hurry and answer please

arlik [135]

It's c (48)

because if you divide all of them its 3

5 0

3 years ago

The distribution of weights for newborn babies is approximately normally distributed with a mean of 7.4 pounds and a standard de

blsea [12.9K]

Answer:

1. 15.87%

2. 6 pounds and 8.8 pounds.

3. 2.28%

4. 50% of newborn babies weigh more than 7.4 pounds.

5. 84%

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 7.4 pounds

Standard Deviation, σ = 0.7 pounds

We are given that the distribution of weights for newborn babies is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

1.Percent of newborn babies weigh more than 8.1 pounds

P(x > 8.1)

$P( x > 8.1) = P( z > \displaystyle\frac{8.1 - 7.4}{0.7}) = P(z > 1)$

$= 1 - P(z \leq 1)$

Calculation the value from standard normal z table, we have,

$P(x > 8.1) = 1 - 0.8413 = 0.1587 = 15.87\%$

15.87% of newborn babies weigh more than 8.1 pounds.

2.The middle 95% of newborn babies weight

Empirical Formula:

Almost all the data lies within three standard deviation from the mean for a normally distributed data.
About 68% of data lies within one standard deviation from the mean.
About 95% of data lies within two standard deviations of the mean.
About 99.7% of data lies within three standard deviation of the mean.

Thus, from empirical formula 95% of newborn babies will lie between

$\mu-2\sigma= 7.4-2(0.7) = 6\\\mu+2\sigma= 7.4+2(0.7)=8.8$

95% of newborn babies will lie between 6 pounds and 8.8 pounds.

3. Percent of newborn babies weigh less than 6 pounds

P(x < 6)

$P( x < 6) = P( z > \displaystyle\frac{6 - 7.4}{0.7}) = P(z < -2)$

Calculation the value from standard normal z table, we have,

$P(x < 6) =0.0228 = 2.28\%$

2.28% of newborn babies weigh less than 6 pounds.

4. 50% of newborn babies weigh more than pounds.

The normal distribution is symmetrical about mean. That is the mean value divide the data in exactly two parts.

Thus, approximately 50% of newborn babies weigh more than 7.4 pounds.

5. Percent of newborn babies weigh between 6.7 and 9.5 pounds

$P(6.7 \leq x \leq 9.5)\\\\ = P(\displaystyle\frac{6.7 - 7.4}{0.7} \leq z \leq \displaystyle\frac{9.5-7.4}{0.7})\\\\ = P(-1 \leq z \leq 3)\\\\= P(z \leq 3) - P(z < -1)\\= 0.9987 -0.1587= 0.84 = 84\%$

84% of newborn babies weigh between 6.7 and 9.5 pounds.

7 0

4 years ago

Bill is able to save $35/week after working part-time and paying his expenses. These two formulas show his weekly savings: f(1)

Umnica [9.8K]

1. We use the recursive formula to make the table of values:
f(1) = 35
f(2) = f(1) + f(2-1) = f(1) + f(1) = 35 + 35 = 70
f(3) = f(1) + f(3-1) = f(1) + f(2) = 35 + 70 = 105
f(4) = f(1) + f(4-1) = f(1) + f(3) = 35 + 105 = 140
f(5) = f(1) + f(5-1) = f(1) + f(4) = 35 + 140 = 175

2. We observe that the pattern is that for each increase of n by 1, the value of f(n) increases by 35. The explicit equation would be that f(n) = 35n. This fits with the description that Bill saves up $35 each week, thus meaning that he adds $35 to the previous week's value.

3. Therefore, the value of f(40) = 35*40 = 1400. This is easier than having to calculate each value from f(1) up to f(39) individually. The answer of 1400 means that Bill will have saved up $1400 after 40 weeks.

4. For the sequence of 5, 6, 8, 11, 15, 20, 26, 33, 41...
The first-order differences between each pair of terms is: 1, 2, 3, 4, 5, 6, 7, 8...since these differences form a linear equation, this sequence can be expressed as a quadratic equation. Since quadratics are functions (they do not have repeating values of the x-coordinate), therefore, this sequence can also be considered a function.

4 0

4 years ago