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2 years ago

11

What’s 6% as a fraction

1 answer:

guapka [62]2 years ago

5 0

Answer: 6/100

Step-by-step explanation:

a percent is out of 100%, to write 6% as a fraction simply but 6/100

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After 1 year, $300 deposited in a savings account with simple interest had earned $12 in

Eddi Din [679]

Answer:

4%

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

A physical education teacher wanted to test whether her county’s students are falling below the national proportion. It is known

Ipatiy [6.2K]

Answer:

(a) The population parameter is students who ran the mile in less than 10 minutes.

(b) The hypothesis being tested is:

H0: p = 0.70

Ha: p < 0.70

(c) The significance level is 0.05.

(d) In order to conduct a one-sample proportion z-test, the following conditions should be met:

The data are a simple random sample from the population of interest.

The population is at least 10 times as large as the sample.

n⋅p ≥ 10 and n⋅(1−p) ≥ 10, where n is the sample size and p is the true population proportion.

n⋅p = 756 * 0.70 ≥ 10

n⋅(1−p) = 756 * (1 - 0.70) ≥ 10

All the conditions are met.

(e) p = 504/756 = 0.67

The test statistic, z = (p - p)/√p(1-p)/n = (0.67 - 0.70)/√0.70(1-0.70)/756 = -2.00

(f) The p-value is 0.0228.

(g) Since the p-value (0.0228) is less than the significance level (0.05), we can reject the null hypothesis.

(h) Therefore, we can conclude that less than 70% of students run one mile in less than 10 minutes.

Step-by-step explanation:

7 0

3 years ago

Find the point (,) on the curve =8 that is closest to the point (3,0). [To do this, first find the distance function between (,)

ELEN [110]

Question:

Find the point (,) on the curve $y = \sqrt x$ that is closest to the point (3,0).

[To do this, first find the distance function between (,) and (3,0) and minimize it.]

Answer:

$(x,y) = (\frac{5}{2},\frac{\sqrt{10}}{2}})$

Step-by-step explanation:

$y = \sqrt x$ can be represented as: $(x,y)$

Substitute $\sqrt x$ for $y$

$(x,y) = (x,\sqrt x)$

So, next:

Calculate the distance between $(x,\sqrt x)$ and $(3,0)$

Distance is calculated as:

$d = \sqrt{(x_1-x_2)^2 + (y_1 - y_2)^2}$

So:

$d = \sqrt{(x-3)^2 + (\sqrt x - 0)^2}$

$d = \sqrt{(x-3)^2 + (\sqrt x)^2}$

Evaluate all exponents

$d = \sqrt{x^2 - 6x +9 + x}$

Rewrite as:

$d = \sqrt{x^2 + x- 6x +9 }$

$d = \sqrt{x^2 - 5x +9 }$

Differentiate using chain rule:

Let

$u = x^2 - 5x +9$

$\frac{du}{dx} = 2x - 5$

So:

$d = \sqrt u$

$d = u^\frac{1}{2}$

$\frac{dd}{du} = \frac{1}{2}u^{-\frac{1}{2}}$

Chain Rule:

$d' = \frac{du}{dx} * \frac{dd}{du}$

$d' = (2x-5) * \frac{1}{2}u^{-\frac{1}{2}}$

$d' = (2x - 5) * \frac{1}{2u^{\frac{1}{2}}}$

$d' = \frac{2x - 5}{2\sqrt u}$

Substitute: $u = x^2 - 5x +9$

$d' = \frac{2x - 5}{2\sqrt{x^2 - 5x + 9}}$

Next, is to minimize (by equating d' to 0)

$\frac{2x - 5}{2\sqrt{x^2 - 5x + 9}} = 0$

Cross Multiply

$2x - 5 = 0$

Solve for x

$2x =5$

$x = \frac{5}{2}$

Substitute $x = \frac{5}{2}$ in $y = \sqrt x$

$y = \sqrt{\frac{5}{2}}$

Split

$y = \frac{\sqrt 5}{\sqrt 2}$

Rationalize

$y = \frac{\sqrt 5}{\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$y = \frac{\sqrt {10}}{\sqrt 4}$

$y = \frac{\sqrt {10}}{2}$

Hence:

$(x,y) = (\frac{5}{2},\frac{\sqrt{10}}{2}})$

3 0

3 years ago

Factor with step by step explanation. n2+40-28

hjlf

N2+40-28=
Combine like terms
n2+40-28=n2+12
Factor out 2
n•2+12=2(n+6)
Therefore the answer is 2(n+6)

5 0

3 years ago

What is the value of y in the equation 5x+2y=20, when x=0.3

Rasek [7]

5x+2y=20

Substitute 0.3 for x

5(0.3)+2y=20

Multiply 5 by 0.3

1.5+2y=20

Subtract 1.5 from both sides

2y=18.5

Divide 2 on both sides

Final Answer: y= 9.25

8 0

3 years ago