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Rus_ich
11 days ago
11

A basketball player makes 90% of her free throws. what is the probability she will miss for the first time on the seventh shot?

Mathematics
1 answer:
Inessa [9K]11 days ago
4 0
The likelihood she will miss on her first attempt is 52.17%.
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Which is the solution of the quadratic equation (4y-3)^2=72 ?
babunello [8412]
Greetings: 
<span>(4y-3)²=72 
4y-3 = </span>±√72
thus, y= (3+√72) /4   or y= (3 - √72) /4
7 0
24 days ago
1-i need to mix 25% mineral spirits to the varnish i am using what ratio of spirit to varnish am i using? 2-if i use 240 ml of v
Svet_ta [9500]

Answer:

(1) The necessary proportion of spirit to varnish stands at 1:3.

(2) A total of 80 ml of mineral spirit is required.

(3) The ratio of the heights to the widths of the tapestries is 3:2.

Step-by-step explanation:

(1) To achieve a mixture containing 25% mineral spirit, the blend consists of 25% spirit and 75% varnish.

Therefore, \frac{spirit}{varnish}=\frac{25}{75}

=\frac{1}{3}

Consequently, the proportion of spirit to varnish is 1:3.

(2) Using 240ml of varnish indicates this is 75% of the entire solution, which means the ratio of varnish to the total solution is 3:4.

Let the total solution quantity be x.

Thus, 240:x=3:4.

⇒\frac{240}{x}=\frac{3}{4}

⇒3x=240*4

⇒x=\frac{240*4}{3}

⇒x=320

This means the total solution amounts to 320 ml.

Now, calculating Spirit = Total solution - Varnish

⇒Spirit = 320ml-240ml

⇒Spirit = 80ml

Therefore, when using 240 ml of varnish, you will require 80 ml of mineral spirit.

(3) The dimensions of Robison's tapestries are uniform at 1.5m in length and 1m in width. Initially, to obtain whole numbers, we multiply these dimensions by 10.

Thus, the dimensions become 15m long and 10m wide.

Now, the proportion of Height to Width is 15:10

⇒\frac{Height}{Width} =\frac{15}{10}

⇒\frac{Height}{Width} =\frac{3}{2}

Thus, the proportion of the heights of the tapestries to their widths is 3:2.

8 0
1 day ago
Which statements describe a residual plot for a line of best fit that is a good model for a scatterplot? Check all that apply.
babunello [8412]

Response: The accurate statements include:-

There are nearly equal quantities of points located above and below the x-axis.

The points are distributed haphazardly without a distinct pattern.

The total number of points matches that of the scatter plot.

Explanation:

  • A residual plot illustrates residuals on the vertical axis against the independent variable on the horizontal axis.

Consequently, the count of points is on par with the scatter plot, and roughly the same amount of points exist above and below the x-axis.

Given the random distribution of the points throughout the plot, it signifies there is no correlation, therefore, the points are scattered randomly without a clear arrangement.


8 0
7 days ago
Read 2 more answers
You are in an airplane 5.7 miles above the ground. What is the measure of BD⌢
AnnZ [9099]

Result:

6.1°; 425.86 m.

Step-by-step breakdown:

The information provided states that the airplane is at an altitude of 5.7 miles above ground level, while the "radius of Earth is about 4000 miles." Thus,

θ = 2 × cos^-1 (a/ (a + b)), where a = 4000 miles, and b = 5.7 miles.

θ = 2 × cos^-1 (4000/ (4000 + 5.7)) = 6.1°.

To calculate the distance in meters:

Change in distance = 6.1° /360° × 2π × 4000 miles = 425.86 meters.

Consequently, BD⌢ measures at 6.1° and the distance corresponding to this section of Earth is 425.86 meters.

6 0
22 days ago
The mass of a colony of bacteria, in grams, is modeled by the function P given by P(t)=2+5tan^−1.(t/2), where t is measured in d
AnnZ [9099]

Answer:

1.21 g/day

Step-by-step explanation:

We start with the fact that

The mass of the bacterial colony (in grams) is described by

P(t)=2+5tan^{-1}(\frac{t}{2})

Where t=Time(in days)

Next, we differentiate with respect to t

P'(t)=5(\frac{1}{1+\frac{t^2}{4}}\times \frac{1}{2})

Using the formula \frac{d(tan^{-1}(x)}{dx}=\frac{1}{1+x^2}

P'(t)=\frac{5}{2}(\frac{4}{4+t^2})

P'(t)=\frac{10}{4+t^2}

We know that P(t)=6

Now, substitute this value

6=2+5tan^{-1}(\frac{t}{2})

5tan^{-1}(\frac{t}{2})=6-2=4

tan^}{-1}(\frac{t}{2})=\frac{4}{5}

\frac{t}{2}=tan(\frac{4}{5})

t=2tan(\frac{4}{5})

Insert the given value of t

P'(2tan\frac{4}{5})=\frac{10}{4+4tan^2(\frac{4}{5})}

P'(2tan\frac{4}{5})=\frac{10}{4}\times \frac{1}{1+tan^2(\frac{4}{5})}

We understand that 1+tan^2\theta=sec^2\theta

Applying the formula

P'(2tan(\frac{4}{5})=\frac{5}{2}\times \frac{1}{sec^2(\frac{4}{5})}

P'(2tan\frac{4}{5})=\frac{5}{2}\times cos^2(\frac{4}{5})

By employing cos^2x=\frac{1}{sec^2x}

P'(2tan\frac{4}{5})=\frac{5}{2}\times (0.696)^2=1.21g/day

Consequently, the instantaneous rate at which the mass of the colony changes is=1.21g/day

7 0
1 month ago
Read 2 more answers
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