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NEW QUESTION: 1
You have an Office 365 subscription that has a site collection with a URL of https://contoso.sharepoint.com.
Users report that they are prevented from uploading documents to the site because of free space issue. You increase the storage quota for the site collection to 1000 MB.
You need to ensure that you receive a warning message when the site collection uses 90 percent of the available space.
What command should you run? To answer, select the appropriate options in the answer area.
NOTE: Each correct selection is worth one point.
Answer:
Explanation:
Explanation
Box1
The Set-SPOSite cmdlet is used to set or updatesone or more properties' values for a site collection.
Box2
The -Identity parameter of the Set-SPOSite cmdlet specifies the URL of the site collection to update.
Box3
The -StorageQuotaWarningLevel parameter of the Set-SPOSite cmdlet specifies the warning level in megabytes of the site collection to warn the site collection administrator that the site is approaching the storage quota.
References: https://technet.microsoft.com/en-us/library/fp161394.aspx
NEW QUESTION: 2
Which CLI command displays the IP configuration of each Data Mover?
A. server_devconfig
B. /sbin/ifconfig
C. server_sysconfig
D. server_ifconfig
Answer: D
NEW QUESTION: 3
The realized mean monthly return on the S&P 500 in the 1990's appears to have been substantially different than the mean return in 2000's. The data indicate that assuming equal population variances is not unreasonable.
u = 1990's population mean return and u = 2000's population mean return. The decision made in this
1 2
hypothesis is (assume at the 10% level):
A. that the t value is significant at 0.1 level.
B. not to reject the null hypothesis.
C. to reject the null hypothesis.
Answer: B
Explanation:
Pooled estimate of variance needs to be computed as follows:
2 2 2
S = [(60 - 1)(5.876) + (60 - 1)(4.986) ] / (60 + 60 - 2) = 29.694.
1/2
Now determine the value of t. t = [(0.7 - 1.8) - 0] / [29.964/60 + 29.964/60] = -1.101. We reject null if t >
1 .658 or t -1.658 (t-value: t(118, 0.05) = 1.658). The t value of -1.101 does not fit the rejection criteria. In other words we do not reject the null hypothesis and the t value is not significant at the 0.1 level.
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