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3 years ago

The first figure is dilated to form the second figure.

Mathematics

1 answer:

Alja [10]3 years ago

5 0

1) The scale factor is 0.25 is true.

Step-by-step explanation:

Step 1:

The first figure is dilated to form the second figure. The shape is a rhombus.

It is given that the side length of the first figure is 5.8 and the side length of the second figure is 1.45.

Step 2:

To calculate the scale factor, we divide the measurement after scaling by the same measurement before scaling. In this case, it is the side length of the rhombus.

The scale factor $= \frac{1.45}{5.8} = \frac{1}{4} = 0.25.$

So the scale factor is 0.25. This is the first option.

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13x-11 +18x15+ 4x+1=180°

BARSIC [14]

Answer:

x = - 80/17, or x = - 4 $\frac{12}{17}$

Step-by-step explanation:

13x-11 +18×15+ 4x+1=180°

<=> (13x + 4x) + (18×15+ 1 - 11)=180°

<=> 17x + 260 =180

<=> 17x = 180 - 260

<=> 17x = - 80

<=> x = - 80/17

<=> x = - 4 $\frac{12}{17}$

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3 years ago

There are 95 candies in the dish.19 of the candies are chocolate. What percent of the candies are left

r-ruslan [8.4K]

Answer:

80%

Step-by-step explanation:

20% of the candies are chocolate, leaving the rest of the 80%

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3 years ago

A standard showerhead in Nathan's house dispenses 11 gallons of water per minute. Nathan changed this showerhead to an energy-sa

Fynjy0 [20]

The new shower head dispenses 3 gallons of water per minute, as shown in the equation y = 3x.

So, 3 x 8 = 24.

Then, to find out how much he saves each day, you would find how much he used before,

11 x 8 = 88

Next we want to see how much Nathan saved, so we subtract the total,

88 - 24 = 64

Nathan saved 64 gallons.

7 0

3 years ago

Read 2 more answers

Suppose that a basketball player can score on a particular shot with probability .3. Use the central limit theorem to find the a

Rom4ik [11]

Answer:

(a) The probability that the number of successes is at most 5 is 0.1379.

(b) The probability that the number of successes is at most 5 is 0.1379.

(d) The probability that the number of successes is at most 11 is 0.9357.

→ All the exact probabilities are more than the approximated probability.

Step-by-step explanation:

Let S = a basketball player scores a shot.

The probability that a basketball player scores a shot is, P (S) = p = 0.30.

The number of sample selected is, n = 25.

The random variable $S\sim Bin(25,0.30)$

According to the central limit theorem if the sample taken from an unknown population is large then the sampling distribution of the sample proportion ( $\hat p$ ) follows a normal distribution.

The mean of the the sampling distribution of the sample proportion is: $E(\hat p)=p=0.30$

The standard deviation of the the sampling distribution of the sample proportion is:

$SD(\hat p)=\sqrt{\frac{ p(1- p)}{n} }=\sqrt{\frac{ 0.30(1-0.30)}{25} }=0.092$

(a)

Compute the probability that the number of successes is at most 5 as follows:

The probability of 5 successes is: $p=\frac{5}{25} =0.20$

$P(\hat p\leq 0.20)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.20-0.30}{0.092} )\\=P(Z\leq -1.087)\\=1-P(Z$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 5 is 0.1379.

The exact probability that the number of successes is at most 5 is:

$P(S\leq 5)={25\choose 5}(0.30)^{5}91-0.30)^{25-5}=0.1935$

The exact probability is more than the approximated probability.

(b)

Compute the probability that the number of successes is at most 7 as follows:

The probability of 5 successes is: $p=\frac{7}{25} =0.28$

$P(\hat p\leq 0.28)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.28-0.30}{0.092} )\\=P(Z\leq -0.2174)\\=1-P(Z$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 7 is 0.4129.

The exact probability that the number of successes is at most 7 is:

$P(S\leq 57)={25\choose 7}(0.30)^{7}91-0.30)^{25-7}=0.5118$

The exact probability is more than the approximated probability.

(c)

Compute the probability that the number of successes is at most 9 as follows:

The probability of 5 successes is: $p=\frac{9}{25} =0.36$

$P(\hat p\leq 0.36)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.36-0.30}{0.092} )\\=P(Z\leq 0.6522)\\=0.7422$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 9 is 0.7422.

The exact probability that the number of successes is at most 9 is:

$P(S\leq 9)={25\choose 9}(0.30)^{9}91-0.30)^{25-9}=0.8106$

The exact probability is more than the approximated probability.

(d)

Compute the probability that the number of successes is at most 11 as follows:

The probability of 5 successes is: $p=\frac{11}{25} =0.44$

$P(\hat p\leq 0.44)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.44-0.30}{0.092} )\\=P(Z\leq 1.522)\\=0.9357$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 11 is 0.9357.

The exact probability that the number of successes is at most 11 is:

$P(S\leq 11)={25\choose 11}(0.30)^{11}91-0.30)^{25-11}=0.9558$

The exact probability is more than the approximated probability.

6 0

3 years ago

What is the equation for x=-3 and y=-7

Juli2301 [7.4K]

Answer:

-7 = -3x + b

Step-by-step explanation:

i'm guessing because you have not provided a visual so i can not determine the B.

8 0

3 years ago