Solve the equation for x.<br> 26x = 156<br> x=

1) What would be another way to name plane M?

garri49 [273]

The correct answer is C plane HJG

4 0

3 years ago

You have a 4 in. X 6in. family picture that you want to resize. You can choose from a 16 in. X 20 in. or an 18 in. X 24 in. Whic

yarga [219]

Answer:

16 in × 20 in

Step-by-step explanation:

Finding the % of the picture that can fit in 16×20 frame

Area of picture= length by width

=4*6=24 in²

Area of frame=Length by width

=16*20=320 in²

% of the picture in the frame will be

(24/320) * 100% =7.5%

2. Finding the % of the picture that can fit in 18 × 24 frame

Area of picture=24 in²

Area of frame =Length by width

=18*24=432 in²

% of the picture in the frame will be;

(24/432)*100%=5.6%

The frame 16 by 20 in will keep 7.5% of the original picture

The frame 18 by 24 in will keep 5.6% of the original picture

Hence you should use the frame of 16 by 20 in because it will keep more of the original picture.

6 0

3 years ago

Although cities encourage carpooling to reduce traffic congestion, most vehicles carry only one person. For example, 64% of vehi

ira [324]

Answer:

a) 0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

b) 0.996 is the probability that more than half of the vehicles carry just one person.

Step-by-step explanation:

We are given the following information:

A) Binomial distribution

We treat vehicle on road with one passenger as a success.

P(success) = 64% = 0.64

Then the number of vehicles follows a binomial distribution, where

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 10

We have to evaluate:

$P(x \geq 6) = P(x =6) +...+ P(x = 10) \\= \binom{10}{6}(0.64)^6(1-0.64)^4 +...+ \binom{10}{10}(0.64)^{10}(1-0.79)^0\\=0.7291$

0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

B) By normal approximation

Sample size, n = 92

p = 0.64

$\mu = np = 92(0.64) = 58.88$

$\sigma = \sqrt{np(1-p)} = \sqrt{92(0.64)(1-0.64)} = 4.60$

We have to evaluate the probability that more than 47 cars carry just one person.

$P(x \geq 47)$

After continuity correction, we will evaluate

$P( x \geq 46.5) = P( z > \displaystyle\frac{46.5 - 58.88}{4.60}) = P(z > -2.6913)$

$= 1 - P(z \leq -2.6913)$

Calculation the value from standard normal z table, we have,

$P(x > 46.5) = 1 - 0.004 = 0.996 = 99.6\%$

0.996 is the probability that more than half out of 92 vehicles carry just one person.

5 0

3 years ago

a vet weighs two puppies the small puppy weighs 4 1/2 pounds the large puppy weighs 4 2/3 times as much as the small puppy how m

Llana [10]

The large puppy weighs 20.97 Pounds. This is because 2/3 of a hundred is 66. You would do 4.5 Times 4.66. This equals 20.97. It can be rounded to 21. Hope this helps!

7 0

4 years ago

Read 2 more answers

For the graph shown, the slope is changed to 1/5, but the y-intercept remains the same. What is the resulting equation for the n

Varvara68 [4.7K]

Answer: C. y=1/5x+1

Step-by-step explanation:

8 0

3 years ago

Solve the equation for x. 26x = 156 x=