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2 years ago

The perimeter of a triangle is 105 cm. If each of the two legs is exactly twice the length of the

Mathematics

1 answer:

Reptile [31]2 years ago

4 0

<em>Greetings from Brasil...</em>

According to the question of the statement, we can conclude that

PQ = 2B

QR = 2B

PR = base = B

Perimeter = P = 105

P = PQ + QR + PR

105 = 2B + 2B + B

B = 21

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Please help me with this math question

earnstyle [38]

80.64
because 16 times 12 is 192
if every tile is 2 feet divide 192 by 2
you should get 96 meaning there are 96 tiles
then take 96 and multiply by it by .84 because every tile is 84 cents
you then get 80.64

6 0

2 years ago

The area of a trapezoid is 144 square inches. if the height is 12 inches, find the length of the median.

deff fn [24]

Answer: 12 inches

Step-by-step explanation: In this problem, since we're asked to find the length of the median, let's use our formula for the area of a trapezoid that involves the median which is shown below.

Area = median · height

We know that the area is 144 and the height is 9 so we can set up the equation 144 = M · 12. Now to solve for <em>m</em>, we divide both sides of the equation by 12 and we find that 12 = M.

So the length of the median of the trapezoid is 12 inches.

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2 years ago

The track team is trying to reduce their time for a relay race. First they reduce their time by 2.1 minutes. Then they are able

konstantin123 [22]

Answer: 15.7 minutes

Step-by-step explanation:

Let x be the time in the beginning (in minutes).

Given: The track team is trying to reduce their time for a relay race.

First they reduce their time by $t_1=$ 2.1 minutes.

Then they are able to reduce that time by $t_2=$ 10

If their final time is 3.96 minutes, then

$x-t_1-t_2=3.6\\\Rightarrow\ x=3.6+t_1+t_2\\\Rightarrow\ x=3.6+2.1+10\\\Rightarrow\ x= 15.7$

Hence, their beginning time was 15.7 minutes.

7 0

2 years ago

a recipe calls for 3 ounces of flour for every 2 ounces of sugar. Find the constant of proportionality.

Ket [755]

3:2

for every 3 ounces of flour, you need 2 ounces of sugar

7 0

2 years ago

Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in

Hitman42 [59]

Answer:

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) We can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval

(3) A survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

Step-by-step explanation:

We are given that a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of Americans who decide to not go to college = 48%

n = sample of American adults = 331

p = population proportion of Americans who decide to not go to

college because they cannot afford it

<em>Here for constructing a 90% confidence interval we have used a One-sample z-test for proportions.</em>

<em />

<u>So, 90% confidence interval for the population proportion, p is ;</u>

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < $\hat p-p$ < $1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

P( $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

<u>90% confidence interval for p</u> = [ $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.48 -1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ , $0.48 +1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ ]

= [0.4348, 0.5252]

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) The interpretation of the above confidence interval is that we can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval.

3) Now, it is given that we wanted the margin of error for the 90% confidence level to be about 1.5%.

So, the margin of error = $Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$