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1 year ago

Find the value of x.

Mathematics

1 answer:

Lina20 [59]1 year ago

4 0

Answer:

Hi,

Step-by-step explanation:

Using Al'Kashi theorem (law of cosines),

x²=26²+19²-2*26*19*cos(31°)

x²= 190,11870690...

$x=\sqrt{190.11870690} =13,7883540...\approx{13.8}$

I hope that this is not a control.

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A bag of chocolates is labeled to contain 0.384 pounds of chocolates. The actual weight of the chocolates is 0.3798 pounds. Are

Marta_Voda [28]

Based on the information given the chocolates stated on the label is 0.0042-ounce lighter.

Weight:

Using this formula

Chocolate weight=Current weight-Actual weight

Where:

Current weight=0.384

Actual weight=0.3798

Let plug in the formula

Chocolate weight=0.384-0.3798

Chocolate weight=0.0042 ounce

In conclusion, the chocolates stated on the label are 0.0042-ounce lighter.

6 0

1 year ago

Masteriza [31]

Answer:

c .....the answer is c.....

Step-by-step explanation:

I hope this helps

8 0

2 years ago

A 12-foot flagpole casts a nine-foot shadow. What is the measure from the top of the shadow to the top of the flag pole?

amm1812

Here is the answer OwO:

A 12-foot flagpole casts a nine-foot shadow. The measure from the top of the shadow to the top of the flag pole is 15.

Here is the explanation/work OwO:

Switch sides.

c^2=12^2+9^2

12^2=144

c^2=144+9^2

9^2=81

c^2=144+81

Add the numbers: 144 + 81 = 225

c^2=225

For x^2 = f(a) the solution is √f(a)

c = 15

Therefore, your answer will be 15.

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Sorry if I am late. Hope this helps!! ^^

5 0

2 years ago

Read 2 more answers

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 201.9-cm and a standard dev

Nesterboy [21]

Answer:

There is a 0.08% probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 201.9-cm and a standard deviation of 2.1-cm. This means that $\mu = 201.9, \sigma = 2.1$ .

For shipment, 9 steel rods are bundled together. Find the probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.

By the Central Limit Theorem, since we are using the mean of the sample, we have to use the standard deviation of the sample in the Z formula. That is:

$s = \frac{\sigma}{\sqrt{n}} = \frac{2.1}{\sqrt{9}} = 0.7$

This probability is 1 subtracted by the pvalue of Z when $X = 204.1$ .

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{204.1 - 201.9}{0.7}$

$Z = 3.14$

$Z = 3.14$ has a pvalue of 0.9992. This means that there is a 1-0.9992 = 0.0008 = 0.08% probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.

3 0

3 years ago

Pls help me ASAP I DONT know what to put. I only have 6 minutes to answer pls help

tester [92]

Answer:

Y= 59* X=20cm?

Step-by-step explanation:

I am not 100% sure but it should be close

6 0

2 years ago