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3 years ago

ANSWER QUICKLY !! How many terms are in “y^2 + 4y - 15” ??? a) two terms b) 3 terms

Mathematics

2 answers:

mestny [16]3 years ago

7 0

Answer:

<h2>b) 3 terms</h2>

Step-by-step explanation:

y² + 4y - 15 = y² + 4y + (-15)

TERMS:

y², 4y, -15

Anton [14]3 years ago

5 0

Answer:3 terms

Step-by-step explanation:

the operation symbols separate the terms

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Find the value of x. help with geometry pls

Ivan

Answer:

Find the value of x:-

To find Y, use Pythagorean theorem:- $c^{2} =a^{2} +b^{2}$

$(2.1)^{2} =y^{2} +(1.4)^{2}$

$2.1^{2}=4.41$

$1.4^{2} =1.96$

$4.41=y^{2} +1.96$

subtract 1.96 from both sides

$2.45=y^{2}$

$y=1.5652$

Now, to find x:-

$x=y+y$

$= 1.5632+1.5652$

$x=3.1 \: ft$

~OAmalOHopeO

6 0

3 years ago

How do i show the work? please help!

asambeis [7]

Tigers: 1 - 2/3 = .3333

Redbirds: 1 - 4/5 = .2

Bulldogs: 1 - 3/8 = .625

Titans: 1 - 1/2 = .5

The Redbirds likelyhood is lowest, therefore, they're least likely to play I'm the championship.

B is the correct answer.

7 0

3 years ago

A coffee packaging plant claims that the mean weight of coffee in its containers is at least 32 ounces. A random sample of 15 co

Luda [366]

Answer:

We conclude that the mean weight of coffee in its containers is at least 32 ounces which means that the data support the claim.

Step-by-step explanation:

We are given that a coffee packaging plant claims that the mean weight of coffee in its containers is at least 32 ounces.

A random sample of 15 containers were weighed and the mean weight was 31.8 ounces with a sample standard deviation of 0.48 ounces.

Let $\mu$ = mean weight of coffee in its containers.

SO, Null Hypothesis, $H_0$ : $\mu \geq$ 32 ounces {means that the mean weight of coffee in its containers is at least 32 ounces}

Alternate Hypothesis, $H_A$ : $\mu$ < 32 ounces {means that the mean weight of coffee in its containers is less than 32 ounces}

The test statistics that would be used here One-sample t-test statistics as we don't know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean weight = 31.8 ounces

s = sample standard deviation = 0.48 ounces

n = sample of containers = 15

So, the test statistics = $\frac{31.8 -32}{\frac{0.48}{\sqrt{15} } }$ ~ $t_1_4$

= -1.614

The value of t test statistics is -1.614.

Now, at 0.01 significance level the t table gives critical value of -2.624 at 14 degree of freedom for left-tailed test.

Since our test statistic is more than the critical value of t as -1.614 > -2.624, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the mean weight of coffee in its containers is at least 32 ounces which means that the data support the claim.

8 0

3 years ago

Use the commutative property for addition to rearrange the algebraic expression 5x-7y-8x+y

BartSMP [9]

You want to use the addition property, so you need to ADD each term. Just change the minuses to pluses, making sure you keep the negative with the number

5x-8x-7y+y =

5x + -8x + -7y +y

5 0

3 years ago

Two studies were done on the same set of data, where study I was a one-sided test and study II was a two-sided test. The p-value

mixer [17]

Answer:

$0.060$

Step-by-step explanation:

In a two tailed test the probability of occurrence is the total area under the critical range of values on both the sides of the curve (negative side and positive side)

Thus, the probability values for a two tailed test as compared to a one tailed test is given by the under given relation -

$p-value = P(Z< -\frac{\alpha }{2} )+P(Z >\frac{\alpha}{2})$ \

Here $P\frac{\alpha}{2} = 0.030$

Substituting the given value in above equation, we get -

probability values for a two tailed test

= $0.030 + 0.030\\= 0.060$

4 0

3 years ago