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2 years ago

The beginning of changing starch to sugar occurs in the stomach.

Mathematics

2 answers:

Black_prince [1.1K]2 years ago

6 0

False. It starts when you first chew it.

borishaifa [10]2 years ago

5 0

False

The beginning of changing starch to sugar occurs in your mouth, by saliva.

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A doctor released the results of clinical trials for a vaccine to prevent a particular disease. In these clinical trials, 200,0

CaHeK987 [17]

Answer:

1) No, there is not enough evidence to support the claim that the vaccine was effective (P-value=0.104) .

The null and alternative hypothesis are:

$H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2< 0$

being the subindex 1 for the experimental group and subindex 2 for the control group.

2) Test statistic z=-1.26

Step-by-step explanation:

This is a hypothesis test for the difference between proportions.

The claim is that the vaccine was effective.

Then, the null and alternative hypothesis are:

$H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2< 0$

The significance level is 0.01.

The sample 1 (experimental group), of size n1=100000 has a proportion of p1=0.0002.

$p_1=X_1/n_1=21/100000=0.0002$

The sample 2 (control group), of size n2=100000 has a proportion of p2=0.0003.

$p_2=X_2/n_2=30/100000=0.0003$

The difference between proportions is (p1-p2)=-0.0001.

$p_d=p_1-p_2=0.0002-0.0003=-0.0001$

The pooled proportion, needed to calculate the standard error, is:

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{21+30}{100000+100000}=\dfrac{51}{200000}=0.00026$

The estimated standard error of the difference between means is computed using the formula:

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.00026*0.99974}{100000}+\dfrac{0.00026*0.99974}{100000}}\\\\\\s_{p1-p2}=\sqrt{0+0.0000000025}=\sqrt{0.0000000051}=0.000071$

Then, we can calculate the z-statistic as:

$z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{-0.00009-0}{0.000071}=\dfrac{-0.00009}{0.000071}=-1.26$

This test is a left-tailed test, so the P-value for this test is calculated as (using a z-table):

$P-value=P(z$

As the P-value (0.104) is bigger than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the vaccine was effective.

6 0

3 years ago

What is the simplified form of (X/3-y/4)/(x/4+y/3) ? (6 points)

Fofino [41]

Option 2: $\frac{4x-3y}{3x+4y}$ is the right answer

Step-by-step explanation:

Given fraction is:

$\frac{\frac{x}{3}-\frac{y}{4}}{\frac{x}{4}+\frac{y}{3}}$

To solve the fraction, Taking LCM in denominator and numerator

$=\frac{\frac{4x-3y}{12} }{\frac{3x+4y}{12}}$

When the numerator and denominator both have fractions, the simplified form is obtained be multiplying the numerator with reciprocal of the denominator. So,

$= \frac{4x-3y}{12} * \frac{12}{3x+4y}\\=\frac{4x-3y}{3x+4y}$

Hence,

Option 2: $\frac{4x-3y}{3x+4y}$ is the right answer

Keywords: fractions, simplification

Learn more about fractions at:

#LearnwithBrainly

4 0

3 years ago

Multiply & Simplify.

kipiarov [429]

2a(7a²-3a+9)
=14a^3 - 6a^2 +18a

answer is D. 14a³-6a²+18a

3 0

3 years ago

Which expression is equivalent to 5y^3/(5y)^-2

Serggg [28]

Answer:

5^3 y^5

125 y^5

Step-by-step explanation:

5y^3/(5y)^-2

Distribute the exponent in the denominator

5y^3/(5 ^-2 y^-2)

A negative exponent in the denominator brings it to the numerator

5y^3 5 ^2 y^2

Combine like terms

5 * 5^2 * y^3 5^2

We know that a^b * a^c = a^(b+c)

5^(1+2) * y^( 3+2)

5^3 y^5

125 y^5

6 0

3 years ago

Dorothy Little purchased a mailing list of 2,000 names and addresses for her mail order business, but after scanning the list sh

DaniilM [7]

Answer:

b) 2.00

Step-by-step explanation:

For each name, there are only two possible outcomes. Either it is authentic, or it is not. The probability of a name being authentic is independent of any other name. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

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

The expected value of the binomial distribution is:

$E(X) = np$