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GREYUIT [131]
3 years ago
11

henry is mixing concrete. to make concrete he needs to mix cement, sand and gravel in the ratio 5 : 4 : 1 by weight. henry needs

to make 110kg of concrete. he has 50kg cement, 55kg of sand and 15kg of gravel. which ingredient is henry short of?
Mathematics
1 answer:
MaRussiya [10]3 years ago
7 0

Answer:

cement by 5 Kg

Step-by-step explanation:

add the parts of the ratio, 5 + 4 + 1 = 10 parts

Divide the quantity of concrete by 10 to find the value of one part of the ratio.

110Kg ÷ 10 = 11Kg ← value of 1 part of ratio , thus

5 parts = 5 × 11Kg = 55Kg ← cement required

4 parts = 4 × 11Kg = 44Kg  ← sand required

1 part = 11Kg ← gravel required

He requires 55Kg of concrete but only has 50Kg.

He requires 44Kg of sand and has 55Kg

He requires 11Kg of gravel and has 15Kg

Thus he is 5Kg short of cement.

You might be interested in
(15) Geometry Help plz
MakcuM [25]
X = measure of angle 1
y = measure of angle 2
z = measure of angle 3
w = measure of angle 4

Focus on the bottom triangle. The three angles add to 180 degrees
(angle 2) + (angle 3) + 116 = 180
y+z+116 = 180
y+z= 180-116
y+z= 64

Since we have the bottom triangle as isosceles, this means that y = z, so
y+z = 64
y+y = 64
2y = 64
y = 64/2
y = 32
making z = 32 as well

Similarly, angle 1 and angle 4 are 32 degrees because the 116 angle is opposite the top left-most angle, and congruent to this angle. In other words, the bottom triangle is a mirror image of the top triangle.

The figure is a rhombus because all four sides are the same length (as shown by the tickmarks)

-----------------------------------------------------------

Answer: 
This figure is a rhombus
All four angles (angle 1 through angle 4) are the same measure. They are each 32 degrees

6 0
3 years ago
Help me out, please.
Assoli18 [71]
Y+4=2/3(x+3)

(-3,-4)(3,0)
1.find the slope
m=y2-y1/x2-x1
m=0-(-4)/3-(-3)
m=4/6
m=2/3

y-(-4)=2/3(x-(-3))
y+4=2/3(x+3) point-slope form

4 0
2 years ago
A researcher conducted a paired sample t-test to determine if advertisements were viewed more in the morning (before noon) or in
Natalija [7]

Answer:

b. No, there was no difference between Morning (M= 32), and Evening (M=40.625), (t [7] = 1.15, p > .05).

Step-by-step explanation:

The critical value for one tailed test is t ∝(7) > 1.895

A one tailed test is performed to test the claim that advertisements were viewed more in the morning (before noon) or in the evening (after 5pm)

The null and alternative hypotheses are

H0: μm = μe   vs     Ha μm > μe

where μm is the mean of the morning and μe is the mean of evening.

The calculated value of t = -1.152587077  which is less than the critical region hence the null hypothesis cannot be rejected .

P(T<=t) one-tail 0.143458126 > 0.05

If two tailed test is performed the critical region is t Critical two-tail 2.364624252

and the calculated t value is  -1.152587077  which again does not lie in the critical region .

Hence μm =  μe or    μm ≤ μe

P(T<=t) two-tail 0.286916252  > 0.025

Therefore

b. No, there was no difference between Morning (M= 32), and Evening (M=40.625), (t [7] = 1.15, p > .05).

Option b gives the best answer.

6 0
3 years ago
Given that a 90% confidence interval for the mean height of all adult males in Idaho measured in inches was [62.532, 76.478]. Us
grigory [225]

Answer:

The confidence interval on this case is given by:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}} (1)

For this case the confidence interval is given by (62.532, 76.478)[/tex]

And we can calculate the mean with this:

\bar X = \frac{62.532+76.478}{2}= 69.505

So then the mean for this case is 69.505

Step-by-step explanation:

Previous concepts

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

\bar X represent the sample mean  

\mu population mean (variable of interest)  

\sigma represent the population standard deviation  

n represent the sample size  

Assuming the X follows a normal distribution  

X \sim N(\mu, \sigma)

The confidence interval on this case is given by:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}} (1)

For this case the confidence interval is given by (62.532, 76.478)[/tex]

And we can calculate the mean with this:

\bar X = \frac{62.532+76.478}{2}= 69.505

So then the mean for this case is 69.505

5 0
3 years ago
The total cost to rent 5 chairs and 3 tables is $27. The total cost to rent 2 chairs and 12 tables is $81. What is the cost to r
kati45 [8]

The cost to rent each chair is $1.5 and cost to rent each table is $6.5

<h3>Applications of systems of linear equations </h3>

From the question, we are to determine the cost to rent each chair and each table

Let c represent chair

and

t represent table

From the given information,

The total cost to rent 5 chairs and 3 tables is $27

That is,

5c + 3t = 27 ------------ (1)

Also,

The total cost to rent 2 chairs and 12 tables is $81

That is,

2c + 12t = 81 ---------- (2)

Now, solve the equations simultaneously

5c + 3t = 27 ------------ (1)

2c + 12t = 81 ---------- (2)

Multiply equation (1) by 2 and multiply equation (2) by 5

2 × [5c + 3t = 27 ]

5 × [2c + 12t = 81 ]

10c + 6t = 54        ------------- (3)

10c + 60t = 405   ------------- (4)

Subtract equation (4) from equation (3)

10c + 6t = 54        

10c + 60t = 405

---------------------------

-54t = -351

t = -351/-54

t = 6.5

Substitute the value of t into equation (2)
2c + 12t = 81

2c + 12(6.5) = 81

2c + 78 = 81

2c = 81 - 78

2c = 3

c = 3/2

c = 1.5

∴ The cost of chair is $1.5 and cost of table is $6.5

Hence, the cost to rent each chair is $1.5 and cost to rent each table is $6.5

Learn more on Solving system of linear equations here: brainly.com/question/13729904

#SPJ1

8 0
1 year ago
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