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Anon25 [30]
3 years ago
15

What is the sum of 103 and h?​

Mathematics
2 answers:
klemol [59]3 years ago
4 0

Answer:

The answer is 103 + h

Step-by-step explanation:

The sum of 103 and h

i.e., 103 + h

Thus, The answer is 103 + h

<u>-TheUnknownScientist</u>

Liula [17]3 years ago
3 0
The answer would be 103 + h If I am wrong I’m sorry
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Please help due in 3 min
andrezito [222]

Answer:

(h*h)(10) is the same as h(10)*h(10). Let's find the value of h(10) first. To do this, replace every x with 10 like so

Step-by-step explanation:

h(x) = 6-x

h(10) = 6-10

h(10) = -4

So,

h(10)*h(10) = (-4)*(-4)

h(10)*h(10) = 16

The final answer is 16

3 0
3 years ago
Read 2 more answers
mya has ballet lessons every sixth day and swimming every fourth day. today she has both lessons. in how many days will mya have
ra1l [238]
To determine when Mya will have both lessons again on the same day, you will list the multiples of each number of days because to show every 4 or 6 days, you will count by 4's and 6's.

When you get to the first number that is the same, that will be the next time she will have both lessons again.  This is called the least common multiple (LCM).

4, 8, 12, 16, 20, ...
6, 12, 18, 24

In 12 days she will have both lessons again.
5 0
4 years ago
The formula for the surface area of a cylinder with radius r and height h is at times twice the product of the
MrMuchimi

Answer:

<em>2\pi \: rh + 2\pi {r}^{2}  = 2(t \times  {r}^{2} )</em>

Step-by-step explanation:

The TSA of the cylinder

2\pi \: rh + 2\pi {r}^{2}

In Mathematics, 'is' is =

Twice means 2( )

Product means ×

So we have

2\pi \: rh + 2\pi {r}^{2}  = 2(t \times  {r}^{2} )

as the answer.

6 0
3 years ago
A report indicated that 37% of adults had received a bogus email intended to steal personal information. Suppose a random sample
fiasKO [112]

Answer:

5.05% probability that no more than 34% had received such an email.

Step-by-step explanation:

We use the binomial approximation to the normal to solve this problem.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

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.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 700, p = 0.37

\mu = E(X) = np = 700*0.37 = 259

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{700*0.37*0.63} = 12.77

In a random sample of 700 adults, what is the probability that no more than 34% had received such an email?

34% is 0.34*700 = 238

So this probability is the pvalue of Z when X = 238.

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

Z = \frac{238 - 259}{12.77}

Z = -1.64

Z = -1.64 has a pvalue of 0.0505

5.05% probability that no more than 34% had received such an email.

7 0
4 years ago
How do you determine the quadratic equation having roots that are real numbers and equal​
zvonat [6]

Answer:

To determine the nature of roots of quadratic equations (in the form ax^2 + bx +c=0) , we need to calculate the discriminant, which is b^2 - 4 a c. When discriminant is greater than zero, the roots are unequal and real. When discriminant is equal to zero, the roots are equal and real.

6 0
3 years ago
Read 2 more answers
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