All categories

3 years ago

Which value for x makes this open sentence true? 5 + 5 ● x = x² + 3²

Mathematics

1 answer:

Angelina_Jolie [31]3 years ago

5 0

The answer your looking for here is 1

You might be interested in

Six students measure the acceleration (in meters per second per second) of object in free fall. The measured values are shown. T

Sunny_sXe [5.5K]

The measured data is
x=[10.56, 9.52, 9.73, 9.80, 9.78, 10.91].

Calculate the mean.
m = (10.56+9.52+9.73+9.80+9.78+10.91)/6 = 10.05

Calculate the deviations from the mean.
k = x - m
= [0.51, -0.53, -0.32, -0.25, -0.27, 0.86]

Calculate the absolute deviation from the mean.
$d= \frac{1}{6} | (0.51-0.53-0.32-0.25-0.27+0.86) | = |1.184\times10^{-15}| \approx 0$

Answer: 0

6 0

4 years ago

NEED HELP ASAP THIS IS A TEST<br><br> will mark brainlest <br><br> NO LINKS

Nataly_w [17]

-371 7/24
From quizlet made from balletskye20 with 16 terms

8 0

3 years ago

What is the answer pleasr

gavmur [86]

For the first one at the top it’s 300 and 575 then going down it’s 17.21, 63.33, and then 184.68

4 0

3 years ago

The answer and how to do this

Vilka [71]

Graphically I guess means graph it.

First graph goes to first problem. Second graph goes to second problem.

5 0

3 years ago

The amount of calories consumed by customers at the Chinese buffet is normally distributed with mean 2885 and standard deviation

aliina [53]

Answer:

a. X~N(2,885, 651)

b. 0.086291

c. 0.00058

d. 3213.10 calories

Step-by-step explanation:

a. -A normal distribution is expressed in the form X~N(mean, standard deviation).

-Let X a random variable denoting the number of calories consumed.

-X is a is a normally distributed random variable with mean 2885 and standard deviation 651.

-This distribution is expressed as X~N(2,885, 651)

b. The probability that less than 2000 calories are consumed is calculated using the formula:

$P(X$

#substitute the given values in the formula to solve for P:

$P(X$

Hence, the probability of consuming less than 2000 calories is 0.08691

c. The proportion of customers consuming more than 5000 calories is calculated as:

$P(X>x)=P(z>\frac{\bar X-\mu}{\sigma})\\\\=P(Z>\frac{5000-2885}{651})\\\\=P(z>3.2488)\\\\=1-0.99942\\\\=0.00058$

Hence, the proportion of customers consuming over 5000 calories is 0.00058

d. The least amount of calories to get the award is calculated as:

1% is equivalent to a z value of 0.50399.

-We equate this to the formula to solve for the mean consumption:

$0.01=P(z>\frac{\bar X-\mu}{\sigma})\\\\=P(z>\frac{\bar X-2885}{651})\\\\\1\%=0.50399 \\\\\frac{\bar X-2885}{651}=0.50399 \\\\\bar X=0.50399\times 651+2885\\\\=3213.09$

Hence, the least amount of calories consumed to qualify for the award is 3213.10 calories.

8 0

3 years ago