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3 years ago

8

Y = -7(-13) 1.-91 2.21 3.91 4.-21

1 answer:

victus00 [196]3 years ago

4 0

The answer is 3 so 91

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2. A rectangle has a perimeter of 100. Its

slava [35]

Answer:

Option H

Step-by-step explanation:

Perimeter of rectangle= 2(length) +2(width)

100= 2(5a -22) +2[½(a +1)]

Expand:

100= 2(5a) +2(-22) +a +1

100= 10a -44 +a +1

Simplify:

100= 11a -43

+43 on both sides:

11a= 100 +43

11a= 143

Divide both sides by 11:

a= 143 ÷11

a= 13

Thus, option H is correct.

8 0

3 years ago

Figures A and B at right are similar. Assuming that figure A is the original figure, find the scale factor and find the lengths

AnnZ [28]

The scale factor is: 1/5.

The missing sides of figure B are: 2.4, 3.6, and 4.

<h3>How to find Scale Factor?</h3>

Scale factor = dimension of new figure / dimension of original figure

The polygons given are shown in the diagram attached below.

Figure A is the original figure

Figure B is the new figure.

x, y, and z has been used to mark the missing sides of figure B.

One dimension of original figure = 15

Corresponding dimension of the new figure = 3

Scale factor = new/original = 3/15

Scale factor = 1/5

To find the missing sides of figure B, multiply each corresponding side lengths of Figure A by the scale factor:

x = 12 × 1/5 = 2.4

y = 18 × 1/5 = 3.6

z = 20 × 1/5 = 4

In summary:

The scale factor is: 1/5.

The missing sides of figure B are: 2.4, 3.6, and 4.

Learn more about scale factor on:

brainly.com/question/2826496

3 0

3 years ago

Tell me which 4 letters are right! A? B? C? D? F? E? I WILL GIVE BRAINLIEST

Helen [10]

Answer:

Everything but A and C

Step-by-step explanation:

5 0

2 years ago

An Airliner has a capacity for 300 passengers. If the company overbook a flight with 320 passengers, What is the probability tha

TEA [102]

Answer:

The probability is $P(X >300 ) = 0.97219$

Step-by-step explanation:

From the question we are told that

The capacity of an Airliner is k = 300 passengers

The sample size n = 320 passengers

The probability the a randomly selected passenger shows up on to the airport

$p = 0.96$

Generally the mean is mathematically represented as

$\mu = n* p$

=> $\mu = 320 * 0.96$

=> $\mu = 307.2$

Generally the standard deviation is

$\sigma = \sqrt{n * p * (1 -p ) }$

=> $\sigma = \sqrt{320 * 0.96 * (1 -0.96 ) }$

=> $\sigma =3.50$

Applying Normal approximation of binomial distribution

Generally the probability that there will not be enough seats to accommodate all passengers is mathematically represented as

$P(X > k ) = P( \frac{ X -\mu }{\sigma } > \frac{k - \mu}{\sigma } )$

Here $\frac{ X -\mu }{\sigma } =Z (The \ standardized \ value \ of \ X )$

=> $P(X >300 ) = P(Z > \frac{300 - 307.2}{3.50} )$

Now applying continuity correction we have

$P(X >300 ) = P(Z > \frac{[300+0.5] - 307.2}{3.50} )$

=> $P(X >300 ) = P(Z > \frac{[300.5] - 307.2}{3.50} )$

=> $P(X >300 ) = P(Z > -1.914 )$

From the z-table

$P(Z > -1.914 ) = 0.97219$

So

$P(X >300 ) = 0.97219$

8 0

3 years ago

With a little care, you can find the median and the quartiles from the histogram. what are these numbers? you can also find the

coldgirl [10]

Complete Question

The histogram from this question is shown on the first uploaded image

Answer:

The first quartile is $1st Q = 19^{th} girl (subject ) = 1 \serving\ of\ fruit\ per\ day$

The median is $Median = 38 ^{th} \ girl (subject) = 2 \serving\ of\ fruit\ per\ day$

The third quartile is $3rd Q = 56^{th} \ girl (subject) = 4 \serving\ of\ fruit\ per\ day$

The mean is $\= x = 2.62$

Step-by-step explanation:

From the question we are told that

The total number of girls is n = 74

Generally the median is mathematically represented as

$Median = \frac{\frac{n}{2} +[ \frac{n}{2} + 1] }{2}$

So

$Median = \frac{\frac{74}{2} +[ \frac{74}{2} + 1] }{2}$

=> $Median = \frac{37 +38 }{2}$

=> $Median =37.5 \ girl$

From the histogram $37.5 \approx 38^{th} \ girl$ fall under 2 fruits per day

Generally the first quartile is mathematically represented as

$1st \ Q = \frac{\frac{n}{4} + [\frac{n}{4} + 1] }{2}$

=> $1st \ Q = \frac{\frac{74}{4} + [\frac{74}{4} + 1] }{2}$

=> $1st \ Q = 19 \ girl$

From the histogram $19^{th}$ girl fall under 1 fruit per day

Generally the third quartile is mathematically represented as

$3rd\ Q = \frac{\frac{n}{2} + n }{2}$

=> $3rd\ Q = \frac{\frac{74}{2} + 74 }{2}$

=> $3rd\ Q = 55.5$

From the histogram $55.5 \approx 56^{th} \ girl$ fall under 4 fruits per day

Generally from the histogram table the frequency [number of subjects (girls) ] and the servings of fruit per day can be represented as

Total

x(servings per day ) 0 1 2 3 4 5 6 7 8

f (number of subjects ) 15 11 15 11 8 5 3 3 3 $\sum f = 74$

xf 0 11 30 33 32 25 18 21 24 $\sum xf = 194$

Generally the mean is mathematically represented as

$\= x = \frac{1}{ \sum f} * [\sum xf]$

=> $\= x = \frac{1}{ 74} *194$

=> $\= x = 2.62$

4 0

3 years ago