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3 years ago

Solve each system of equation Y=-3x+5 5x-4y=-3

1 answer:

6 0

Y = -3x + 5
5x - 4y = -3

   5x - 4(-3x + 5) = -3
5x + 4(3x) - 4(5) = -3
   5x + 12x - 20 = -3
   17x - 20 = -3
   + 20 + 20
      17x = 17
   17    17
      x = 1
      y = -3x + 5
      y = -3(1) + 5
      y = -3 + 5
      y = 2
      (x, y) = (1, 2)

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The HCP prescribes methotrexate 7.5 mg PO weekly, in 3 divides doses for a child with rheumatoid arthritis whose body surface ar

dexar [7]

Answer:

1.5mg

Step-by-step explanation:

From the question, we are told that the HCP prescribed 7.5 mg of PO weekly

The therapeutic dosage is given in the question as 5 - 15 mg/m² weekly.

The child's body surface area is given = 0.6m²

The mg of PO that the nurse should administer in each of the three doses given weekly is calculated as

7.5mg/ 5mg/m²

= 1.5 mg of PO

4 0

3 years ago

Which expressions are equivalent to 12x - 6?

iren2701 [21]

<h3>Answer</h3>

B and E

<h3>Explanation</h3>

A) It's wrong because

$- 6 \times (2x - 1) = - 12x + 6$

B) It's right because

$6 \times (2x - 1) = 12x - 6$

C) It's wrong because

$6x \times (2 - 1) = 6x \times 1 = 6x$

D) It's wrong because

$- 6x \times (2x - 1) = - 12 {x}^{2} + 6$

E) It's right because

$- 6 \times ( - 2x + 1) = 12x - 6$

F) It's wrong because

$6 \times ( - 2x + 1) = - 12x + 6$

8 0

3 years ago

A pole that is 2.8 m tall casts a shadow that is 1.67 m long. At the same time, a nearby building casts a shadow that is 50.75 m

Veseljchak [2.6K]

Answer:

Explanation:

3.2

1.14

44.75

Cross Mutliply:

44.75

⋅

3.2

1.14

Divide:

143.2

1.14

You get 125.614 rounded to the nearest meter is 126

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

What goes where. JdhshshHshxjancixd

valentina_108 [34]

Answer:

$\boxed { \frac{7}{5}y } \to \: \boxed {y+ \frac{2}{5}y }$

$\boxed { \frac{3}{5}y } \to \: \boxed {y - \frac{2}{5}y }$

Step-by-step explanation:

We need to simplify

$y + \frac{2}{5}y$

We collect LCM to get;

$\frac{5y + 2y}{5} = \frac{7y}{5}$

Therefore:

$\boxed { \frac{7}{5}y } \to \: \boxed {y+ \frac{2}{5}y }$

Also we need to simplify:

$y - \frac{2}{5}y$

We collect LCM to get;

$y - \frac{2}{5}y = \frac{5y - 2y}{5} = \frac{3}{5} y$

Therefore

$\boxed { \frac{3}{5}y } \to \: \boxed {y - \frac{2}{5}y }$

5 0

3 years ago

A study of long-distance phone calls made from General Electric's corporate headquarters in Fairfield, Connecticut, revealed the

Jet001 [13]

Answer:

a) 0.4332 = 43.32% of the calls last between 3.6 and 4.2 minutes

b) 0.0668 = 6.68% of the calls last more than 4.2 minutes

c) 0.0666 = 6.66% of the calls last between 4.2 and 5 minutes

d) 0.9330 = 93.30% of the calls last between 3 and 5 minutes

e) They last at least 4.3 minutes

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 3.6, \sigma = 0.4$

(a) What fraction of the calls last between 3.6 and 4.2 minutes?

This is the pvalue of Z when X = 4.2 subtracted by the pvalue of Z when X = 3.6.

X = 4.2

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

$Z = \frac{4.2 - 3.6}{0.4}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

X = 3.6

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

$Z = \frac{3.6 - 3.6}{0.4}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5

0.9332 - 0.5 = 0.4332

0.4332 = 43.32% of the calls last between 3.6 and 4.2 minutes

(b) What fraction of the calls last more than 4.2 minutes?

This is 1 subtracted by the pvalue of Z when X = 4.2. So

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

$Z = \frac{4.2 - 3.6}{0.4}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

1 - 0.9332 = 0.0668

0.0668 = 6.68% of the calls last more than 4.2 minutes

(c) What fraction of the calls last between 4.2 and 5 minutes?

This is the pvalue of Z when X = 5 subtracted by the pvalue of Z when X = 4.2. So

X = 5

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

$Z = \frac{5 - 3.6}{0.4}$

$Z = 3.5$

$Z = 3.5$ has a pvalue of 0.9998

X = 4.2

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

$Z = \frac{4.2 - 3.6}{0.4}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

0.9998 - 0.9332 = 0.0666

0.0666 = 6.66% of the calls last between 4.2 and 5 minutes

(d) What fraction of the calls last between 3 and 5 minutes?

This is the pvalue of Z when X = 5 subtracted by the pvalue of Z when X = 3.

X = 5

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

$Z = \frac{5 - 3.6}{0.4}$

$Z = 3.5$

$Z = 3.5$ has a pvalue of 0.9998

X = 3

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

$Z = \frac{3 - 3.6}{0.4}$

$Z = -1.5$

$Z = -1.5$ has a pvalue of 0.0668

0.9998 - 0.0668 = 0.9330

0.9330 = 93.30% of the calls last between 3 and 5 minutes

(e) As part of her report to the president, the director of communications would like to report the length of the longest (in duration) 4% of the calls. What is this time?

At least X minutes

X is the 100-4 = 96th percentile, which is found when Z has a pvalue of 0.96. So X when Z = 1.75.

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

$1.75 = \frac{X - 3.6}{0.4}$

$X - 3.6 = 0.4*1.75$

$X = 4.3$

They last at least 4.3 minutes

7 0

3 years ago