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Marat540 [252]
3 years ago
7

The space allowed for the mascot on a​ school's Web page is 75 pixels wide by 60 pixels high. Its digital image is 500 pixels wi

de by 400 pixels high. What is the largest image of the mascot that will fit on the Web​ page?
Mathematics
1 answer:
anzhelika [568]3 years ago
6 0

Answer:

It will be 60 pixels high and 75 pixels wide

Step-by-step explanation:

\frac{400}{60}=\frac{500}{x}

As

  • The space allowed for the mascot on a​ school's Web page is 75 pixels wide by 60 pixels high
  • Its digital image is 500 pixels wide by 400 pixels high

So, the expression becomes

\frac{400}{60}=\frac{500}{x}

\mathrm{Apply\:fraction\:cross\:multiply:\:if\:}\frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c

400x=60\cdot \:500

400x=30000

\mathrm{Divide\:both\:sides\:by\:}400

\frac{400x}{400}=\frac{30000}{400}

\mathrm{Simplify}

x=75

Therefore, it will be 60 pixels high and 75 pixels wide

Keywords: pixel, ratio

Learn more about ratio from brainly.com/question/4287633

#learnwithBrainly

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Find the supplement of 80º (only answer with a number, not the word<br> degrees)
Nana76 [90]

Answer:

100°

Step-by-step explanation:

two angles are supplement .

if their sum is equal to 180°

supplement angle of  80° + x = 180°

 180° - 80° = 100° answer  

8 0
3 years ago
Test the hypothesis using the​ P-value approach. Be sure to verify the requirements of the test. Upper H 0​: pequals0.6 versus U
kow [346]

Answer:

Step-by-step explanation:

From the given information:

The null and the alternative hypothesis can be well written as:

H_o:P=0.6

H_1:P>0.6

Given that:

n = 200

x = 135

Alpha ∝ = 0.05 level of significance

Then;

⇒ n \times p\times (1-P)

= 200 × 0.6 × (1 -0.6)

= 200 × 0.6 × 0.4

= 48 ≥ 10

The sample proportion \hat P = \dfrac{x}{n}

= \dfrac{135}{200}

= 0.675

The test statistics Z = \dfrac{\hat P - P}{\sqrt{ \dfrac{P(1-P)}{n} }}

Z = \dfrac{0.675 - 0.6}{\sqrt{ \dfrac{0.6 \times 0.4}{200} }}

Z = \dfrac{0.075}{\sqrt{ \dfrac{0.24}{200} }}

Z = 2.165

The P-value = P(Z > 2.165)

= 1 - P(Z < 2.165)

From the z tables

= 1 - 0.9848

= 0.0152

Reject the null hypothesis since P-Value is lesser than alpha. ( i.e. 0.0152 < 0.05).

Thus, there is enough evidence to conclude that the value of the population proportion is greater than 0.6

7 0
3 years ago
The table of values represents an exponential function f(x).
Ostrovityanka [42]

Answer:

-3.98 (nearest hundredth)

Step-by-step explanation:

The average rate of change of function f(x) over the interval a ≤ x ≤ b is given by:

\dfrac{f(b)-f(a)}{b-a}

Given interval:  -2 ≤ x ≤ 2

\implies a = -2

\implies b = 2

\implies f(a) = f(-2)=16

\implies f(b) =f(2)= \dfrac{1}{16}

Substituting the values into the equation:

\begin{aligned}\implies \textsf{rate of change} & =\dfrac{\frac{1}{16}-16}{2-(-2)}\\\\ & = \dfrac{-\frac{255}{16}}{4}\\\\ & = -\dfrac{255}{64}\\\\ & = -3.98\: \sf (nearest\:hundredth)\end{aligned}

7 0
2 years ago
3/4 of what number is 8 more than 25
Ainat [17]

Answer:

3/4 of 44 is 8 more than 25

Step-by-step explanation:

8 more than 25, is 33.

33 is 3/4 of 44.

because 33 divides by 3 will be 11, then times 4, you will get 44.

7 0
3 years ago
Assume that females have pulse rates that are normally distributed with a mean of μ=73.0 beats per minute and a standard deviati
Gennadij [26K]

Answer:

a. the probability that her pulse rate is less than 76 beats per minute is 0.5948

b. If 25 adult females are randomly​ selected,  the probability that they have pulse rates with a mean less than 76 beats per minute is 0.8849

c.   D. Since the original population has a normal​ distribution, the distribution of sample means is a normal distribution for any sample size.

Step-by-step explanation:

Given that:

Mean μ =73.0

Standard deviation σ =12.5

a. If 1 adult female is randomly​ selected, find the probability that her pulse rate is less than 76 beats per minute.

Let X represent the random variable that is normally distributed with a mean of 73.0 beats per minute and a standard deviation of 12.5 beats per minute.

Then : X \sim N ( μ = 73.0 , σ = 12.5)

The probability that her pulse rate is less than 76 beats per minute can be computed as:

P(X < 76) = P(\dfrac{X-\mu}{\sigma}< \dfrac{X-\mu}{\sigma})

P(X < 76) = P(\dfrac{76-\mu}{\sigma}< \dfrac{76-73}{12.5})

P(X < 76) = P(Z< \dfrac{3}{12.5})

P(X < 76) = P(Z< 0.24)

From the standard normal distribution tables,

P(X < 76) = 0.5948

Therefore , the probability that her pulse rate is less than 76 beats per minute is 0.5948

b.  If 25 adult females are randomly​ selected, find the probability that they have pulse rates with a mean less than 76 beats per minute.

now; we have a sample size n = 25

The probability can now be calculated as follows:

P(\overline X < 76) = P(\dfrac{\overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{ \overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}})

P( \overline X < 76) = P(\dfrac{76-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{76-73}{\dfrac{12.5}{\sqrt{25}}})

P( \overline X < 76) = P(Z< \dfrac{3}{\dfrac{12.5}{5}})

P( \overline X < 76) = P(Z< 1.2)

From the standard normal distribution tables,

P(\overline X < 76) = 0.8849

c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

In order to determine the probability in part (b);  the  normal distribution is perfect to be used here even when the sample size does not exceed 30.

Therefore option D is correct.

Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

5 0
2 years ago
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