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Mkey [24]
3 years ago
15

Can you help me with this problem?​

Mathematics
1 answer:
Ede4ka [16]3 years ago
6 0

Answer:

Di ko po alam grade 4palang po ako

Step-by-step explanation:

I'm so sorry po

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If two lines are perpendicular, their slopes are negative reciprocals.<br> A. True<br> B.False
Ainat [17]
A. If the slopes of two lines are negative then the reciprocals, of the lines are perpendicular.
8 0
3 years ago
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A bag contains 6 red apples and 5 yellow apples. 3 apples are selected at random. Find the probability of selecting 1 red apple
tester [92]

Answer:  The required probability of selecting 1 red apple and 2 yellow apples is 36.36%.

Step-by-step explanation:  We are given that a bag contains 6 red apples and 5 yellow apples out of which 3 apples are selected at random.

We are to find the probability of selecting 1 red apple and 2 yellow apples.

Let S denote the sample space for selecting 3 apples from the bag and let A denote the event of selecting 1 red apple and 2 yellow apples.

Then, we have

n(S)=^{6+5}C_3=^{11}C_3=\dfrac{11!}{3!(11-3)!}=\dfrac{11\times10\times9\times8!}{3\times2\times1\times8!}=165,\\\\\\n(A)\\\\\\=^6C_1\times^5C_2\\\\\\=\dfrac{6!}{1!(6-1)!}\times\dfrac{5!}{2!(5-2)!}\\\\\\=\dfrac{6\times5!}{1\times5!}\times\dfrac{5\times4\times3!}{2\times1\times3!}\\\\\\=6\times5\times2\\\\=60.

Therefore, the probability of event A is given by

P(A)=\dfrac{n(A)}{n(S)}=\dfrac{60}{165}=\dfrac{4}{11}\times100\%=36.36\%.

Thus, the required probability of selecting 1 red apple and 2 yellow apples is 36.36%.

4 0
3 years ago
A certain rare form of cancer occurs in 37 children in a​ million, so its probability is 0.000037. In the city of Normalville th
sladkih [1.3K]

Answer:

Mean = 2.7

In a group of 74090 we would expect about 3 (rounding to nearest whole number) children with the rare form of cancer.

Step-by-step explanation:

We are given that the rate of cancer in children is 37 children in 1 million. So the probability of cancer in a child is P(C) = 0.000037

Poisson distribution is used to approximate the number of cases of diseases and we have to find what will be the mean number of cases for 74,090.

In simple words we have to find the expected number of children with cancer in a group of 74,090 children.

The mean value of expected value can be obtained by multiplying the probability with the sample size. So, in this case multiplying probability of child having a cancer with total group size will give us the expected or mean number of children in the group with cancer.

Mean = E(x) = P(C) * Group size

Mean = 0.000037 x 74090

Mean = 2.7

This means in a group of 74090 we would expect about 3 (rounding to nearest whole number) children with the rare form of cancer.

5 0
3 years ago
Please help mee! <br> no links :(
Cerrena [4.2K]

Answer:

a=0.62, b=7/9, c=10/9

Step-by-step explanation:

Brainliest would be appreciated

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In a random sample of cars driven at low altitudes, of them exceeded a standard of grams of particulate pollution per gallon of
Orlov [11]

Complete question is;

In a random sample of 370 cars driven at low altitudes, 43 of them exceeded a standard of 10 grams of particulate pollution per gallon of fuel consumed. In an independent random sample of 80 cars driven at high altitudes, 23 of them exceeded the standard. Can you conclude that the proportion of high-altitude vehicles exceeding the standard is greater than the proportion of low-altitude vehicles exceeding the standard at an level of significance? Group of answer choices

Answer:

Yes we can conclude that there is enough evidence to support the claim that the proportion of high-altitude vehicles exceeding the standard is greater than the proportion of low-altitude vehicles exceeding the standard (P-value = 0.00005).

Step-by-step explanation:

This is a hypothesis test for the difference between the proportions.

The claim is that the proportion of high-altitude vehicles exceeding the standard is greater than the proportion of low-altitude vehicles exceeding the standard.

Then, the null and alternative hypothesis are:

H0 ; π1 - π2 = 0

H1 ; π1 - π2 < 0

The significance level would be established in 0.01.

The random sample 1 (low altitudes), of size n1 = 370 has a proportion of;

p1 = x1/n1

p1 = 43/370

p1 = 0.116

The random sample 2 (high altitudes), of size n2 = 80 has a proportion of;

p2 = x2/n2

p2 = 23/80

p2 = 0.288

The difference between proportions is pd = (p1-p2);

pd = p1 - p2 = 0.116 - 0.288

pd = -0.171

The pooled proportion, we need to calculate the standard error, is:

p = (x1 + x2)/(n1 + n2)

p = (43 + 23)/(370 + 80)

p = 66/450

p = 0.147

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

S_(p1-p2) = √[((p(1 - p)/n1) + ((p(1 - p)/n2)]

1 - p = 1 - 0.147 = 0.853

Thus;

S_(p1-p2) = √[((0.147 × 0.853)/370) + ((0.147 × 0.853)/80)]

S_(p1-p2) = 0.044

Now, we can use the formula for z-statistics as;

z = (pd - (π1 - π2))/S_(p1-p2)

z = (-0.171 - 0)/0.044

z = -3.89

Using z-distribution table, we have the p-value = 0.00005

Since the P-value of (0.00005) is smaller than the significance level (0.01), then the effect is significant.

We conclude that The null hypothesis is rejected.

Thus, there is enough evidence to support the claim that the proportion of high-altitude vehicles exceeding the standard is greater than the proportion of low-altitude vehicles exceeding the standard.

6 0
3 years ago
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