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2 years ago

How do you solve the equation 2/7m-1/7=3/14

1 answer:

8 0

Add 1/7 to both sides to get 2/7m=3/14+1/7 find common denominater (14) so multipy 1/7 by 2 to get 2/14 than add that to 3/14 to get 5/14 than take 2/7 and 5/14 turn into a decimals than divide the 2/7 on both sides to get m=?

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Which rule describes the composition of transformations that maps pre-image ABCD to final image A"B"C"D"? Reflection across the

NeTakaya

Answer:

Your answer is the first point

A reflection across the x-axis and then a transformation of -6 and 1.

7 0

2 years ago

A glass ball is packed in the box shown. Which expression represents the volume of foam padding needed to fill the space in the

zhuklara [117]

Answer: (27- 4/3 pi) r^3

Step-by-step explanation: 1. Volume of a cube: V= a^3, V= 3^3 , V=27

2. Volume of a sphere: V=4/3 pi r^3 ......

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2 years ago

Find the number a such that the line x = a bisects the area under the curve y = 1/x2 for 1 ≤ x ≤ 4. 8 5 (b) find the number b s

IceJOKER [234]

a. We're looking for $a$ such that

$\displaystyle\int_1^a\frac{\mathrm dx}{x^2}=\int_a^4\frac{\mathrm dx}{x^2}$

$-\dfrac1x\bigg|_{x=1}^{x=a}=-\dfrac1x\bigg|_{x=a}^{x=4}$

$1-\dfrac1a=\dfrac1a-\dfrac14$

$\dfrac54=\dfrac2a\implies\boxed{a=\dfrac85}$

b. Integrating with respect to $y$ will make things easier.

$y=\dfrac1{x^2}\implies x=\dfrac1{\sqrt y}$

(where we take the positive square root because we know $x>0$ )

Now we want to find $b$ such that

$\displaystyle\int_0^b\frac{\mathrm dy}{\sqrt y}=\int_b^1\frac{\mathrm dy}{\sqrt y}$

$2\sqrt y\bigg|_{y=0}^{y=b}=2\sqrt y\bigg|_{y=b}^{b=1}$

$2\sqrt b=2-2\sqrt b$

$4\sqrt b=2\implies\boxed{b=\dfrac14}$

3 0

2 years ago

Casey has a small business making dessert baskets. She estimates that her fixed weekly costs for rent, electricity, and salaries

Daniel [21]

500=2.5x+200

x=the amount of desserts.

500-200=2.5x
300=2.5x
x=120
C is the answer.

4 0

2 years ago

Read 2 more answers

The sales data for January and February of a frozen yogurt shop are approximately normal. The mean daily sales for January was $

VashaNatasha [74]

Answer:

January had a higher z-score for sales on the 15th, and the value of that z-score was of 0.5.

Step-by-step explanation:

z-score:

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

January:

The mean daily sales for January was $300 with a standard deviation of $20. On the 15th of January, the shop sold $310 of yogurt. This means, respectively, that $\mu = 300, \sigma = 20, X = 310$ . So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{310 - 300}{20}$

$Z = 0.5$

February:

The mean daily sales for February was $320 with a standard deviation of $50. On the 15th of February, the shop sold $340 of yogurt. This means, respectively, that $\mu = 320, \sigma = 50, X = 340$ . So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{340 - 320}{50}$

$Z = 0.4$

January had a higher z-score for sales on the 15th, and the value of that z-score was of 0.5.

3 0

2 years ago