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2 years ago

3 A buoy is 30 feet from the shore. You swim 3/5 the way to the buoy. How much farther do you have to swim to reach the buoy?

Mathematics

1 answer:

disa [49]2 years ago

8 0

Answer:

12 feet

Step-by-step explanation:

30 divided by 5= 6

6 times 3= how far youve already swam(18 feet)

6 times 2= how far you have left(12 feet)

You might be interested in

Write an inequality to compare the integers -5 and -6

Dovator [93]

-5 > -6

I think this is what it would be because -5 is greater than (>) -6

Hope this is right and it helps :)

5 0

2 years ago

The table below shows the amount paid for different numbers of items. Determine if this relationship forms a direct variation. V

Elden [556K]

<h2>Answer/Step-by-step explanation:</h2>

Direct variation occurs when a variable varies directly with another variable. That is, as the x-variable increases, the y-variable also increases.

The ratio of between y-variable and x-variable would be constant.

Direct variation can be represented by the equation, $y = xk$ , where k is a constant. Thus,

$\frac{y}{x} = k$

From the table given, it seems, as x increases, y also increases. Let's find out if there is a constant of proportionality (k).

Thus, ratio of y to x, $\frac{0.50}{1} = 0.5$

k = 0.5.

If the given table of values has a direct variation relationship, then, plugging in the values of any (x, y), into $\frac{y}{x} = k$ , should give us the same constant if proportionality.

Let's check:

When x = 2, and y = 1:

$\frac{y}{x} = k$ ,

$\frac{1}{2} = 0.5$ ,

When x = 3, y = 1.5:

$\frac{1.5}{3} = 0.5$ ,

When x = 5, y = 2.50:

$\frac{2.5}{5} = 0.5$ ,

The constant of proportionality is the same. Therefore, the relationship forms a direct variation.

5 0

2 years ago

Line segments XY and ZY are tangent to circle O. Circle O is shown. Triangle X Y Z has points X and Z on the circle. Lines X Y a

pickupchik [31]

Answer:

The correct option is;

An isosceles triangle

Step-by-step explanation:

The parameters given are;

Line segments XY and ZY are tangents to circle O.

ΔXYZ has points X and Z on the circle.

XY and ZY are tangents that intersect at point Y outside the circle'

Therefore, given the circle has center O, then;

OY ≅ OZ = Radius of circle with center O

∠OXY ≅ ∠OZY = 90° (The radius line from the circle center to the tangent point is perpendicular)

OY ≅ OY (Reflexive property)

ΔOYX and ΔOYZ are right triangles (Triangle with one angle = 90°

ΔOYX ≅ ΔOYZ (Hypotenuse Leg, HL, rule of congruency)

Therefore;

XY = ZY (Corresponding Parts of Congruent Triangles are Congruent CPCTC)

Triangle XYZ is an isosceles triangle (Triangle with two equal sides).

4 0

2 years ago

A parabolic satellite dish reflects signals to the dish’s focal point. An antenna designer analyzed signals transmitted to a sat

ivolga24 [154]

Answer:

$c=\frac{16}{39}$

Step-by-step explanation:

The probability density function is :

$f(x)=c(1-\frac{1}{16}x^{2})$

With 0 < x < 3

To be a valid probability density function :

$\int\limits^b_a {f(x)} \,dx=1$

Where a < x < b

And also

f(x) ≥ 0 for a < x < b

Applying this to the probability density function of the exercise :

$\int\limits^3_0 {c(1-\frac{1}{16}x^{2})} \, dx=1$

$c\int\limits^3_0 {(1-\frac{1}{16}x^{2})} \, dx=1$

$c(3-\frac{1}{16}\frac{3^{3}}{3})=1$

$c(\frac{39}{16})=1$

$c=\frac{16}{39}$

We can verify by replacing ''c'' in the original probability density function and integrating :

$\int\limits^3_0 {\frac{16}{39}(1-\frac{1}{16}x^{2})} \, dx=$

$=\frac{16}{39}.(3)-\frac{1}{39}.(\frac{3^{3}}{3})=\frac{16}{13}-\frac{3}{13}=\frac{13}{13}=1$

Also, f(x) ≥ 0 for 0 < x < 3

4 0

2 years ago

Exhibit 11-2 We are interested in determining whether the variances of the sales at two music stores (A and B) are equal. A samp

Fiesta28 [93]

Answer:

Option A , should not be rejected

Step-by-step explanation:

From the question we are told that:

Sample 1 $n=25$

Standard deviation 1 $\sigma_1=30$

Sample 2 $n=16$

Standard deviation 2 $\sigma_2=20$

Level of confidence $\alpha=95\%=>0.95$

Generally The Hypothesis are given as

$H_0:\sigma_1^2=\sigma_1^2$

$H_1:\sigma_1^2\neq \sigma_1^2$

Generally the equation for test Statistics is mathematically given by

$T=\frac{\sigma_1^2}{\sigma_2^2}$

$T=\frac{30^2}{20^2}$

$T=2.25$

Therefore

Critical Value

$X=(\alpha,df_1)$

$X'=(\alpha,df_2)$

Where

$df=n-1$

Therefore

$X=(0.95,24)$

$X'=(0.95,15)$

From Table

$T_{Critical}=T_x,T_{x'}$

$T_{Critical}=0.41,2.701$

Therefore

$T_x$

Hence,We fail to reject the Null Hypothesis $H_0$

Option A , should not be rejected

3 0

1 year ago

3 A buoy is 30 feet from the shore. You swim 3/5 the way to the buoy. How much farther do you have to swim to reach the buoy?​

3 A buoy is 30 feet from the shore. You swim 3/5 the way to the buoy. How much farther do you have to swim to reach the buoy?