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3 years ago

A pitcher struck out 11 out of 40 batters. Write an equivalent decimal.

Mathematics

1 answer:

Komok [63]3 years ago

5 0

Answer:

0.275

Step-by-step explanation:

We want to write 11 out of 40 batters as a decimal.

We can write 11 out of 40 as a fraction to get:

$\frac{11}{40}$

Note that:

$\frac{11}{40} = \frac{1}{4} \times \frac{11}{10}$

This implies that:

$\frac{11}{40} = 0.25 \times 1.1$

This gives us

$\frac{11}{40 } = 0.275$

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THE NUBER OF PEOPLE CONTACTED

Elis [28]

Answer:

option c, x = 3

Step-by-step explanation:

Given:

f(x) = 27

f(x) = $3^{x}$

Since both equal f(x) they must be equivalent as well,

$3^{x} = 27\\$

Apply exponent rule, 3 x 3 x 3 = 27

hence x = 3

4 0

2 years ago

Juan wants to know the cross-sectional area of a circular pipe. He measures the diameter which he finds, to the nearest millimet

Tpy6a [65]

This question is not correctly written.

Correct Question

Juan wants to know the cross- sectional area of a circular pipe. He measures the diameter which he finds, to the nearest millimeter, to be 5 centimeters. To find the area of the circle, Juan uses the formula A=〖πr〗^2 where A is the area of the circle and r is its radius. He uses 3.14 for π.

a) What value does Juan get for the area of the circle? Make sure you include your units.

b) Michelle found the area of a circle as 78.5 〖in〗^2. She used 3.14 for π. What is the radius of the circle?

Answer:

a)The Area of the circle is 1962.5 millimeters ²

b) The radius of the circle is

5inches( in)

Step-by-step explanation:

a) The formula for the Area of a circle = πr²

The diameter of the circle was measured by Juan to the nearest millimeters = 5 centimeters

The first step is to convert 5cm to millimeters

1 centimetre = 10 millimeters

5 centimetres =

Cross multiply

= 5 × 10 millimeters

Therefore, 5 centimetres = 50 millimeters.

Since the Diameter of the circle = 50 millimeters,

Radius is calculated as Diameter ÷ 2

Therefore , the radius of the circle = 50 millimeters ÷ 2

= 25 millimeters.

The second step is to find the Area of the circle.

Area of the circle = πr²

Juan used 3.14 for π

Therefore , Area of the circle is calculated as:

= 3.14 ×(25millimeters)²

= 3.14 × 625 millimeters²

= 1962.5 millimeter ²

The Area of the circle is 1962.5 millimeters²

b) Michelle found the area of a circle as 78.5 〖in〗^2. She used 3.14 for π. What is the radius of the circle?

The formula to calculate the Area of a Circle is given as:

A( Area of a circle) = πr²

Since we are asked to find the radius of the circle in Part B, we can derive the formula to find the radius of a circle

A = πr²

r² = A ÷ π

r = √A÷ π

Therefore , the formula to find the Radius of a circle is given as :

r = √A ÷ π

Already given in the question are the following values :

Area of the circle = 78.5inches²

π = Michelle used 3.14

The radius of the circle

= r = √A ÷ π

r = √78.5 in² ÷ 3.14

r = √ 25

r = 5inches( in)

Therefore the radius of the circle is

5inches(in).

5 0

3 years ago

A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l

Katena32 [7]

Answer:

(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is 0.1271.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.

(e) The time is 5.65 minutes.

Step-by-step explanation:

We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.

Let X = the length of the calls, in minutes.

So, X ~ Normal( $\mu=4.5,\sigma^{2} =0.70^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean time = 4.5 minutes

$\sigma$ = standard deviation = 0.7 minutes

(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X $\leq$ 4.50 min)

P(X < 5.30 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{5.30-4.5}{0.7}$ ) = P(Z < 1.14) = 0.8729

P(X $\leq$ 4.50 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.5-4.5}{0.7}$ ) = P(Z $\leq$ 0) = 0.50

The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)

P(X > 5.30 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{5.30-4.5}{0.7}$ ) = P(Z > 1.14) = 1 - P(Z $\leq$ 1.14)

= 1 - 0.8729 = 0.1271

The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 5.30 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 5.30 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{5.30-4.5}{0.7}$ ) = P(Z $\leq$ 1.14) = 0.8729

The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 4.00 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 4.00 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.0-4.5}{0.7}$ ) = P(Z $\leq$ -0.71) = 1 - P(Z < 0.71)

= 1 - 0.7612 = 0.2388

The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = 0.745.

(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;

P(X > x) = 0.05 {where x is the required time}

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-4.5}{0.7}$ ) = 0.05