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3 years ago

What is the solution of the equation? 2^5√(x + 6)^3 + 3 = 19

1 answer:

3 0

For the solution I isolated the variable by dividing each side by factors that don't contain the variable.
x = -6 + ∛2 / 2
x ≈ -5.37003947

Hope this helps!! :3
(If not, sorry)

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Please help mee

NemiM [27]

(2^6)^x=1
(4)^x=1
(4)^0=1
1=1
X=0

(5^0)^x=1
(1)^x=1
x=(-∞,∞)

5 0

2 years ago

....pls help me :((((((

Rina8888 [55]

Answer:

y= -7

Step-by-step explanation:

So first distribute the 8

4=8y+32-4y

now combine like terms (the -4y+8y)

4=32+4y

Subtract 32 on both sides to isolate 4y

-28=4y

now divide 4 on both sides to isolate Y

-7=y

so y= -7

5 0

2 years ago

At the instant that a cake is removed from the oven, the temperature of the cake is 350°F. After 10 minutes, the cake's temperat

BARSIC [14]

Once the cake is removed from the oven, it has been exposed to the room temperature of 70°F. After 10 minutes, the cake's temperature decreased to 200°F, which is 150°F cooler than it initially was (350°F). As for the what is asked, 90°F is 260°F cooler than its original temperature (350°F).

This problem can be expressed in a ratio:
10mins:150°F = N:260°F
where N is how long it will take for cake to cool to 90°F

150°FxN=10x260
150N=2600
N=2600/150
N=17.33
N≈20 minutes

Thus, it takes approximately 20 minutes for the cake to cool to 90°F

8 0

3 years ago

The scores of students on the ACT college entrance exam in a recent year had the normal distribution with mean  =18.6 and stand

Maurinko [17]

Answer:

a) 33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) 0.39% probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are 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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 18.6, \sigma = 5.9$

a) What is the probability that a single student randomly chosen from all those taking the test scores 21 or higher?

This is 1 subtracted by the pvalue of Z when X = 21. So

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

$Z = \frac{21 - 18.6}{5.4}$

$Z = 0.44$

$Z = 0.44$ has a pvalue of 0.67

1 - 0.67 = 0.33

33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) The average score of the 76 students at Northside High who took the test was x =20.4. What is the probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher?

Now we have $n = 76, s = \frac{5.9}{\sqrt{76}} = 0.6768$