If you can get 4 bottles of fruit juice for 10 dollars how much would 1 bottle cost

What is the answer if u know it?

Molodets [167]

Answer:

minimum point

(5,4)(-2,4)

3 0

2 years ago

Read 2 more answers

Solve 1)3x-7=35? 2)22=7-3a 3)-d+7=3 4)2x+23=49 5)-c+2=5

astra-53 [7]

$1)\ 3x-7=35\ \ \ |add\ 7\ to\ both\ sides\\\\3x=42\ \ \ \ |divide\ both\ sides\ by\ 3\\\\\boxed{x=14}\\-------------------------\\2)\ 22=7-3a\\\\7-3a=22\ \ \ \ |subtract\ 7\ from\ both\ sides\\\\-3a=15\ \ \ \ \ |divide\ both\ sides\ by\ (-3)\\\\\boxed{a=-5}\\--------------------------$

$3)\ -d+7=3\ \ \ \ \ |subtract\ 7\ from\ both\ sides\\\\-d=-4\\\\\boxed{d=4}\\---------------------------\\\\4)\ 2x+23=49\ \ \ \ \ \ |subtract\ 23\ from\ both\ sides\\\\2x=26\ \ \ \ \ \ |divide\ both\ sides\ by\ 2\\\\\boxed{x=13}\\---------------------------\\5)\ -c+2=5\ \ \ \ \ \ |subtract\ 2\ from\ both\ sides\\\\-c=3\\\\\boxed{c=-3}$

7 0

2 years ago

Read 2 more answers

In the figure, BE||CD. Solve for x. Show your work.

marin [14]

Answer:

Answer is x = 15

Answer is given below with explanations.

Step-by-step explanation:

$in \: the \: figure \: BE || CD \\ by \: thales \: theorem \\ \frac{AE}{ED} = \frac{AB}{BC} \\ on \: substituting \: the \: values \: from \: the \: figure \\ \frac{9}{6} = \frac{x}{10} \\ \frac{3}{2} = \frac{x}{10} \\ x = \frac{3 \times 10}{2} \\ x = \frac{30}{2} \\ x =15 \\$

HAVE A NICE DAY!

THANKS FOR GIVING ME THE OPPORTUNITY TO ANSWER YOUR QUESTION.

3 0

2 years ago

Horizontal pieces can be produced at a rate of 200 per labor hour. a carpenter uses 500 horizontal pieces and 22 vertical pieces

ira [324]

The number of vertical pieces will be 28. Then the correct option is A.
The number of horizontal pieces will be 600. Then the correct option is B.

<h3>What is decision-making?</h3>

The process of making choice is by identifying the correct decision, gathering information, and assessing alternative solutions.

Horizontal pieces can be produced at a rate of 200 per labor hour.

A carpenter uses 500 horizontal pieces and 22 vertical pieces while making racks for an hour.

You have 900 horizontal pieces and 50 vertical pieces in stock right now.

For the next hour, you assign Juan to rack making, Debbie to trim making, and both mike and stan to horizontal piece production.

1. Then at the end of the next hour, the number of vertical pieces available in stock will be

→ 50 - 22

→ 28

Thus, the number of verticle pieces will be 28.

2. Then at the end of the next hour, then the number of horizontal pieces available in stock will be

→ 900 + 200 - 500

→ 600

More about the decision-making link is given below.

brainly.com/question/3369578

#SPJ1

7 0

1 year ago

An industrial sewing machine uses ball bearings that are targeted to have a diameter of 0.75 inch. The lower and upper specifica

Natalija [7]

Answer:

(a) Probability that a ball bearing is between the target and the actual mean is 0.2734.

(b) Probability that a ball bearing is between the lower specification limit and the target is 0.226.

(c) Probability that a ball bearing is above the upper specification limit is 0.0401.

(d) Probability that a ball bearing is below the lower specification limit is 0.0006.

Step-by-step explanation:

We are given that an industrial sewing machine uses ball bearings that are targeted to have a diameter of 0.75 inch. The lower and upper specification limits under which the ball bearings can operate are 0.74 inch and 0.76 inch, respectively.

Past experience has indicated that the actual diameter of the ball bearings is approximately normally distributed, with a mean of 0.753 inch and a standard deviation of 0.004 inch.

Let X = diameter of the ball bearings

SO, X ~ Normal( $\mu=0.753,\sigma^{2} =0.004^{2}$ )

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

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

where, $\mu$ = population mean = 0.753 inch

$\sigma$ = standard deviation = 0.004 inch

(a) Probability that a ball bearing is between the target and the actual mean is given by = P(0.75 < X < 0.753) = P(X < 0.753 inch) - P(X $\leq$ 0.75 inch)

P(X < 0.753) = P( $\frac{X-\mu}{\sigma} } }$ < $\frac{0.753-0.753}{0.004} } }$ ) = P(Z < 0) = 0.50

P(X $\leq$ 0.75) = P( $\frac{X-\mu}{\sigma} } }$ $\leq$ $\frac{0.75-0.753}{0.004} } }$ ) = P(Z $\leq$ -0.75) = 1 - P(Z < 0.75)

= 1 - 0.7734 = 0.2266

The above probability is calculated by looking at the value of x = 0 and x = 0.75 in the z table which has an area of 0.50 and 0.7734 respectively.

Therefore, P(0.75 inch < X < 0.753 inch) = 0.50 - 0.2266 = 0.2734.

(b) Probability that a ball bearing is between the lower specification limit and the target is given by = P(0.74 < X < 0.75) = P(X < 0.75 inch) - P(X $\leq$ 0.74 inch)

P(X < 0.75) = P( $\frac{X-\mu}{\sigma} } }$ < $\frac{0.75-0.753}{0.004} } }$ ) = P(Z < -0.75) = 1 - P(Z $\leq$ 0.75)

= 1 - 0.7734 = 0.2266

P(X $\leq$ 0.74) = P( $\frac{X-\mu}{\sigma} } }$ $\leq$ $\frac{0.74-0.753}{0.004} } }$ ) = P(Z $\leq$ -3.25) = 1 - P(Z < 3.25)

= 1 - 0.9994 = 0.0006

The above probability is calculated by looking at the value of x = 0.75 and x = 3.25 in the z table which has an area of 0.7734 and 0.9994 respectively.

Therefore, P(0.74 inch < X < 0.75 inch) = 0.2266 - 0.0006 = 0.226.

(c) Probability that a ball bearing is above the upper specification limit is given by = P(X > 0.76 inch)

P(X > 0.76) = P( $\frac{X-\mu}{\sigma} } }$ > $\frac{0.76-0.753}{0.004} } }$ ) = P(Z > -1.75) = 1 - P(Z $\leq$ 1.75)

= 1 - 0.95994 = 0.0401

The above probability is calculated by looking at the value of x = 1.75 in the z table which has an area of 0.95994.

(d) Probability that a ball bearing is below the lower specification limit is given by = P(X < 0.74 inch)

P(X < 0.74) = P( $\frac{X-\mu}{\sigma} } }$ < $\frac{0.74-0.753}{0.004} } }$ ) = P(Z < -3.25) = 1 - P(Z $\leq$ 3.25)

= 1 - 0.9994 = 0.0006

The above probability is calculated by looking at the value of x = 3.25 in the z table which has an area of 0.9994.

8 0

3 years ago